(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\).

Substitute into the original equation: \((y+1) + 2(y+1)^{1/3} + 1 = 0\).

This simplifies to \(2(y+1)^{1/3} = -y - 2\).

Cube both sides: \(8(y+1) = -(y+2)^3\).

Expand and simplify: \(8y + 8 = -y^3 - 6y^2 - 12y - 8\).

Rearrange to get the cubic equation: \(y^3 + 6y^2 + 20y + 16 = 0\).

(b) Use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Given \(\alpha + \beta + \gamma = 0\), \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\), and \(\alpha\beta\gamma = -1\), calculate \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\).

(c) Use the result from (a) or a valid formula for the sum of cubes: \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\).

(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\). Substitute into the original equation:

\((y+1)^3 + 2(y+1) + 1 = 0\)

\(\Rightarrow 2(y+1) = -y - 2\)

\(8(y+1) = -(y+2)^3\)

\(\Rightarrow 8y + 8 = -y^3 - 6y^2 - 12y - 8\)

\(y^3 + 6y^2 + 20y + 16 = 0\)

(b) \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\)

(c) \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\)

(a) The determinant of the matrix \(\begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}\) is calculated as:

\(\left| \begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 1 & \gamma \\ \gamma & 1 \end{array} \right| - \alpha \cdot \left| \begin{array}{cc} \alpha & \gamma \\ \beta & 1 \end{array} \right| + \beta \cdot \left| \begin{array}{cc} \alpha & 1 \\ \beta & \gamma \end{array} \right|\)

\(= 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Since the matrix is singular, the determinant is zero:

\(1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 0\)

Given \(\alpha\beta\gamma = 1\), we have:

\(\alpha^2 + \beta^2 + \gamma^2 = 1 + 2(1) = 3\)

(b) We use the identity:

\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)

Given \(\alpha^2 + \beta^2 + \gamma^2 = 3\), we have:

\(3 = b^2 - 2c\)

Also, given \(\alpha^3 + \beta^3 + \gamma^3 = 3\), we use:

\(\alpha^3 + \beta^3 + \gamma^3 = -b(3) - c(-b) + 3\)

\(3 = -3b + cb + 3\)

Solving these simultaneous equations, we find:

\(c = 3, \ b = 3\)

(a) By Vieta's formulas, the sum \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{b}{a} = -\frac{1}{2}p\).

(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factorized as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). Using Vieta's formulas, \(\alpha + \beta + \gamma = -\frac{1}{2}\) and \(\alpha\beta\gamma = -\frac{5}{2}\). Thus, the value is \(-\frac{5}{2} \times -\frac{1}{2} = \frac{5}{4}\).

(c) The roots \(\alpha\beta, \beta\gamma, \alpha\gamma\) imply a new cubic equation. Using Vieta's formulas, the sum of the roots is \(\alpha\beta + \beta\gamma + \alpha\gamma = -\frac{1}{2}p\), the sum of the products of the roots taken two at a time is \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \frac{5}{4}\), and the product of the roots is \((\alpha\beta)(\beta\gamma)(\alpha\gamma) = (\alpha\beta\gamma)^2 = \left(-\frac{5}{2}\right)^2 = \frac{25}{4}\). The cubic equation is \(4z^3 + 2pz^2 - 5z - 25 = 0\).

(d) Given \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), we use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the known values, \(\frac{1}{3} = \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}p\right)\). Solving for \(p\), we get \(\frac{1}{3} = \frac{1}{4} + p\), leading to \(p = \frac{1}{12}\).

(a) By Vieta's formulas, the sum of the products of the roots taken two at a time is given by \(-\frac{b}{a}\) for the equation \(ax^3 + bx^2 + cx + d = 0\). Here, \(a = 2\), \(b = 1\), so \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{1}{2}p\).

(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factored as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). From Vieta's formulas, \(\alpha\beta\gamma = \frac{5}{2}\) and \(\alpha + \beta + \gamma = -\frac{1}{2}\). Therefore, \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \frac{5}{2} \times -\frac{1}{2} = -\frac{5}{4}\).

(c) The roots of the new cubic equation are \(\alpha\beta, \beta\gamma, \alpha\gamma\). The sum of these roots is \(\alpha\beta + \beta\gamma + \alpha\gamma = -\frac{1}{2}p\), the sum of the products of the roots taken two at a time is \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = -\frac{5}{4}\), and the product of the roots is \((\alpha\beta)(\beta\gamma)(\alpha\gamma) = (\alpha\beta\gamma)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}\). The cubic equation is \(4z^3 + 2pz^2 - 5z - 25 = 0\).

(d) Given \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), we use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the known values, \(\frac{1}{3} = \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}p\right)\). Solving for \(p\), \(\frac{1}{3} = \frac{1}{4} + p\), so \(p = \frac{1}{12}\).

(a) Let \(y = \alpha^2 + 1\). Then \(\alpha^2 = y - 1\). Substitute into the original equation:

\((y - 1)^3 + 2(y - 1)^2 + 3(y - 1) + 1 = 0\)

Expanding and simplifying gives:

\(y^3 - y^2 + 4y - 5 = 0\)

(b) Using the result from part (a), we have:

\((\alpha^2 + 1)^2 + (\beta^2 + 1)^2 + (\gamma^2 + 1)^2 = 1 - 2(4) = -7\)

(c) Again using the result from part (a):

\((\alpha^2 + 1)^3 + (\beta^2 + 1)^3 + (\gamma^2 + 1)^3 = -7 - 4 + 5(3) = 4\)

(a) Let \(y = x^{-3}\), then \(x = y^{-\frac{1}{3}}\). Substitute into the original equation:

\(2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0\)

Multiply through by \(y\) to eliminate the fractions:

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(125y = (6y - 2)^3\)

Expanding and simplifying gives:

\(216y^3 - 216y^2 - 53y - 8 = 0\)

(b) Using the identity \(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1\) and \(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\frac{53}{216}\), we find:

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = (\alpha^{-3} + \beta^{-3} + \gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3})\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\frac{53}{216}\right)\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \frac{161}{108}\)

(c) Using the equation \(216S_9 = 216S_6 + 53S_3 + 24\) and substituting the values from parts (a) and (b):

\(S_9 = \frac{399}{216} = \frac{133}{72}\)

(a) By Vieta's formulas, we have:

\(\alpha + \beta + \gamma = -5\)

\(\alpha\beta + \beta\gamma + \gamma\alpha = 10\)

\(\alpha\beta\gamma = 2\)

We use the identity:

\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)

\(\alpha^2 + \beta^2 + \gamma^2 = (-5)^2 - 2(10) = 25 - 20 = 5\)

(b) To show the matrix is singular, we find its determinant:

\(\begin{vmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{vmatrix} = 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Substituting the known values:

\(1 - 5 + 2(2) = 0\)

Thus, the determinant is zero, and the matrix is singular.

(a) From the roots \(\alpha, \alpha, \gamma\), we have:

\(2\alpha + \gamma = -b\)

\(\alpha^2 + 2\alpha\gamma = 0\)

Solving, \(\alpha = -2\gamma\) leading to \(-4\gamma + \gamma = -b\), so \(\gamma = \frac{1}{3}b\) and \(\alpha = -\frac{2}{3}b\).

\(\alpha^2\gamma = -d\) leading to \(\frac{4}{27}b^3 = -d\) leading to \(4b^3 + 27d = 0\).

(b) Given \(2\alpha^2 + \gamma^2 = 3b\), substitute for \(\alpha\) and \(\gamma\):

\(3b = b^2\) leading to \(b = 3\).

Substitute back to find \(d\):

\(d = -4\).

(a) The value of \(S_1\) is given by the sum of the roots: \(S_1 = \alpha + \beta + \gamma = 2\).

To find \(S_2\), use the relation \(S_2 = S_1^2 - 2(0) = 4\).

(b)(i) The recurrence relation is \(S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n\).

(b)(ii) Using the recurrence relation, \(S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1 = 4\).

(c) Substitute \(x = 2 - y\) into the cubic equation: \(2(2-y)^3 - 4(2-y)^2 + 3 = 0\). Simplifying gives \(2y^3 - 8y^2 + 8y - 3 = 0\).

(d) The value of \(\frac{1}{\alpha + \beta} + \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha}\) is \(\frac{8}{3}\).

Given \(\alpha + \beta + \gamma = 3\), we have \(b = - (\alpha + \beta + \gamma) = -3\).

Using the formula for the sum of squares: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\),

\(5 = 3^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Solving gives \(\alpha\beta + \beta\gamma + \gamma\alpha = 1\), so \(c = 2\).

Using the formula for the sum of cubes: \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\),

\(6 = 3(\alpha\beta\gamma) + 3(5 - 1)\).

Solving gives \(\alpha\beta\gamma = 1\), so \(d = 1\).

The equation is \(x^3 - 3x^2 + 2x + 1 = 0\).

Answer:

1. A cubic equation with roots \(\alpha^3,\beta^3,\gamma^3\) is \(x^3+3x^2+(3+c^3)x+1=0\).
2. \(\alpha^6+\beta^6+\gamma^6=3-2c^3\).
3. The real value is \(c=\sqrt[3]{2}\).

1. Let \(y=x^3\), so \(x=y^{1/3}\). Since \(x\) satisfies \(x^3+cx+1=0\),

   \(y+cy^{1/3}+1=0\).

   Hence

   \(cy^{1/3}=-(y+1)\).

   Cubing both sides gives

   \(c^3y=-(y+1)^3\).

   So

   \((y+1)^3+c^3y=0\).

   Expanding,

   \(y^3+3y^2+3y+1+c^3y=0\),

   and therefore

   \(y^3+3y^2+(3+c^3)y+1=0\).

   Thus a cubic equation whose roots are \(\alpha^3,\beta^3,\gamma^3\) is \(x^3+3x^2+(3+c^3)x+1=0\).

2. From part (a), the roots \(\alpha^3,\beta^3,\gamma^3\) have sum \(-3\) and sum of pairwise products \(3+c^3\). Therefore

   \(\alpha^3+\beta^3+\gamma^3=-3\),

   and

   \(\alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3=3+c^3\).

   Now use \(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)\), with \(a=\alpha^3\), \(b=\beta^3\), and \(c=\gamma^3\). Then

   \(\alpha^6+\beta^6+\gamma^6=(\alpha^3+\beta^3+\gamma^3)^2-2(\alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3)\).

   Substituting the values above,

   \(\alpha^6+\beta^6+\gamma^6=(-3)^2-2(3+c^3)=9-6-2c^3=3-2c^3\).

3. Let \(u=\alpha^3\), \(v=\beta^3\), and \(w=\gamma^3\). The matrix is

   \(\begin{pmatrix}1&u&v\\u&1&w\\v&w&1\end{pmatrix}\).

   Its determinant is

   \(1-u^2-v^2-w^2+2uvw\).

   Here

   \(u^2+v^2+w^2=\alpha^6+\beta^6+\gamma^6=3-2c^3\).

   Also, from \(x^3+cx+1=0\), the product of the roots is

   \(\alpha\beta\gamma=-1\),

   so

   \(uvw=\alpha^3\beta^3\gamma^3=(\alpha\beta\gamma)^3=(-1)^3=-1\).

   Therefore the determinant is

   \(1-(3-2c^3)+2(-1)=2c^3-4\).

   For the matrix to be singular, this determinant must be zero:

   \(2c^3-4=0\).

   Thus

   \(c^3=2\),

   so the real value of \(c\) is

   \(c=\sqrt[3]{2}\).

Answer: (a) A cubic equation with roots \(\alpha^2,\beta^2,\gamma^2\) is \(36y^3-(p^2+36)y^2+(10p+9)y-25=0\).

(b)(i) \(p=-6\).

(b)(ii) \(\alpha^3+\beta^3+\gamma^3=5\).

For \(6x^3+px^2-3x-5=0\), with roots \(\alpha,\beta,\gamma\), the standard root sums are

\(\alpha+\beta+\gamma=-\frac{p}{6}\), \(\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{1}{2}\), and \(\alpha\beta\gamma=\frac{5}{6}\).

(a) Let the new roots be \(\alpha^2,\beta^2,\gamma^2\). Their sum is

\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=\left(-\frac{p}{6}\right)^2-2\left(-\frac{1}{2}\right)=\frac{p^2+36}{36}\).

The sum of their pairwise products is

\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)\).

So

\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=\left(-\frac{1}{2}\right)^2-2\left(\frac{5}{6}\right)\left(-\frac{p}{6}\right)=\frac{1}{4}+\frac{5p}{18}=\frac{10p+9}{36}\).

Their product is

\(\alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=\left(\frac{5}{6}\right)^2=\frac{25}{36}\).

Hence a cubic with roots \(\alpha^2,\beta^2,\gamma^2\) is

\(y^3-\frac{p^2+36}{36}y^2+\frac{10p+9}{36}y-\frac{25}{36}=0\).

Multiplying by \(36\),

\(36y^3-(p^2+36)y^2+(10p+9)y-25=0\).

(b)(i) From the result above,

\(\alpha^2+\beta^2+\gamma^2=\frac{p^2+36}{36}\).

Also

\(2(\alpha+\beta+\gamma)=2\left(-\frac{p}{6}\right)=-\frac{p}{3}\).

Using \(\alpha^2+\beta^2+\gamma^2=2(\alpha+\beta+\gamma)\),

\(\frac{p^2+36}{36}=-\frac{p}{3}\).

Multiplying by \(36\),

\(p^2+36=-12p\), so \(p^2+12p+36=0\).

Therefore \((p+6)^2=0\), giving \(p=-6\).

(b)(ii) When \(p=-6\), the original equation is

\(6x^3-6x^2-3x-5=0\).

So each root \(r\) satisfies

\(6r^3=6r^2+3r+5\).

Applying this to \(\alpha,\beta,\gamma\) and adding gives

\(6(\alpha^3+\beta^3+\gamma^3)=6(\alpha^2+\beta^2+\gamma^2)+3(\alpha+\beta+\gamma)+15\).

For \(p=-6\), \(\alpha+\beta+\gamma=1\), and the given condition gives \(\alpha^2+\beta^2+\gamma^2=2\). Therefore

\(6(\alpha^3+\beta^3+\gamma^3)=6(2)+3(1)+15=30\).

Hence

\(\alpha^3+\beta^3+\gamma^3=5\).

Answer: (i) The equation with roots \(\alpha^3,\beta^3,\gamma^3\) is \(y^3+3y^2+2y+1=0\), so \(\alpha^3+\beta^3+\gamma^3=-3\).

(ii) \(S_{-3}=-2\).

(iii) \(S_6=5\) and \(S_9=-12\).

(i) Let \(y=x^3\). Since \(x^3-x+1=0\), we have

\(y-x+1=0\), so \(x=y+1\).

But \(x=y^{1/3}\), hence

\(y^{1/3}=y+1\).

Cubing both sides gives

\(y=(y+1)^3=y^3+3y^2+3y+1\),

so

\(y^3+3y^2+2y+1=0\).

Therefore \(\alpha^3,\beta^3,\gamma^3\) are the roots of this equation. By Vieta's formula,

\(\alpha^3+\beta^3+\gamma^3=-3\).

(ii) Put \(A=\alpha^3\), \(B=\beta^3\), \(C=\gamma^3\). For the cubic \(y^3+3y^2+2y+1=0\),

\(A+B+C=-3\), \(AB+BC+CA=2\), and \(ABC=-1\).

Hence

\(S_{-3}=\alpha^{-3}+\beta^{-3}+\gamma^{-3}=\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\dfrac{AB+BC+CA}{ABC}=\dfrac{2}{-1}=-2\).

(iii) Since \(A=\alpha^3\), \(B=\beta^3\), and \(C=\gamma^3\),

\(S_6=\alpha^6+\beta^6+\gamma^6=A^2+B^2+C^2\).

Therefore

\(S_6=(A+B+C)^2-2(AB+BC+CA)=(-3)^2-2(2)=5\),

as required.

For \(S_9\), use the equation satisfied by \(A,B,C\):

\(y^3+3y^2+2y+1=0\), so \(y^3=-3y^2-2y-1\).

Summing this over \(A,B,C\),

\(S_9=-3S_6-2S_3-3\).

Since \(S_6=5\) and \(S_3=A+B+C=-3\),

\(S_9=-3(5)-2(-3)-3=-12\).

Answer: (i) \(c=5\), \(d=12\).

(ii) \(\alpha^2+\beta^2+\gamma^2=b^2-10\).

(iii) \(b^3-15b+126=0\).

(iv) \(b=-6\).

For \(x^3+bx^2+cx+d=0\), Vieta's formulae give

\(\alpha+\beta+\gamma=-b\),

\(\alpha\beta+\beta\gamma+\gamma\alpha=c\),

\(\alpha\beta\gamma=-d\).

(i) Since

\(\displaystyle \frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac{c}{-12}\),

and this is \(-\dfrac5{12}\), we get \(c=5\).

Also \(\alpha\beta\gamma=-12\), so \(-d=-12\), and hence \(d=12\).

(ii)

\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\).

Therefore

\(\alpha^2+\beta^2+\gamma^2=(-b)^2-2(5)=b^2-10\).

(iii) Each root satisfies

\(x^3+bx^2+5x+12=0\).

Summing this equation over \(x=\alpha,\beta,\gamma\),

\(\alpha^3+\beta^3+\gamma^3+b(\alpha^2+\beta^2+\gamma^2)+5(\alpha+\beta+\gamma)+36=0\).

Substitute \(\alpha^3+\beta^3+\gamma^3=90\), \(\alpha^2+\beta^2+\gamma^2=b^2-10\), and \(\alpha+\beta+\gamma=-b\):

\(90+b(b^2-10)-5b+36=0\).

Hence

\(b^3-15b+126=0\),

as required.

(iv) The equation \(y^3-15y+126=0\) has real coefficients. Since \(3+i\sqrt{12}\) is a root, \(3-i\sqrt{12}\) is also a root.

The sum of the roots is \(0\), because there is no \(y^2\) term. Therefore the third root is

\(-\big((3+i\sqrt{12})+(3-i\sqrt{12})\big)=-6\).

Since \(b\) is the real root of \(b^3-15b+126=0\), we have \(b=-6\).

Answer: (i) The roots \(\alpha,\beta,\gamma\) satisfy

\(\alpha+\beta+\gamma=-2,\quad \alpha\beta+\beta\gamma+\gamma\alpha=1,\quad \alpha\beta\gamma=-7\)

by Vieta’s formulae for \(x^{3}+2x^{2}+x+7=0\).

Now let

\(y=\dfrac{\alpha}{\beta\gamma}\).

Since \(\beta\gamma=\dfrac{-7}{\alpha}\), we get \(y=\dfrac{\alpha^{2}}{-7}\), so \(\alpha^{2}=-7y\). Thus each root \(\alpha,\beta,\gamma\) gives a corresponding value of \(y\), and the resulting cubic in \(y\) has roots \(\dfrac{\alpha}{\beta\gamma},\dfrac{\beta}{\gamma\alpha},\dfrac{\gamma}{\alpha\beta}\).

Using the standard transformation for the reciprocal-type quantities, these three values are the roots of \(49y^{3}+14y^{2}-27y+7=0\).

(ii) Let

\(u=\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}},\quad v=\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}},\quad w=\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}.\)

Then

\(u=\left(\dfrac{\alpha}{\beta\gamma}\right)^{2},\ v=\left(\dfrac{\beta}{\gamma\alpha}\right)^{2},\ w=\left(\dfrac{\gamma}{\alpha\beta}\right)^{2}.\)

So \(u,v,w\) are the squares of the roots from part (i). If those roots are \(r_1,r_2,r_3\), then from \(49y^{3}+14y^{2}-27y+7=0\),

\(r_1+r_2+r_3=-\dfrac{14}{49}=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=\dfrac{-27}{49}.\)

Therefore

\(u+v+w=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)\)

\(=\left(-\dfrac{2}{7}\right)^2-2\left(-\dfrac{27}{49}\right)=\dfrac{4}{49}+\dfrac{54}{49}=\dfrac{58}{49}.\)

(iii) Let

\(S_3=\dfrac{\alpha^{3}}{\beta^{3}\gamma^{3}}+\dfrac{\beta^{3}}{\gamma^{3}\alpha^{3}}+\dfrac{\gamma^{3}}{\alpha^{3}\beta^{3}}.\)

Using \(u,v,w\) as above, we have \(S_3=u^{3/2}+v^{3/2}+w^{3/2}\), but it is easier to use the roots \(r_1,r_2,r_3\) from part (i):

\(S_3=r_1^{3}+r_2^{3}+r_3^{3}.\)

From the cubic in part (i), the elementary symmetric sums are

\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49},\quad r_1r_2r_3=-\dfrac{1}{7}.\)

Now use

\(r_1^{3}+r_2^{3}+r_3^{3}=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)

So

\(S_3=\left(-\dfrac{2}{7}\right)^3-3\left(-\dfrac{2}{7}\right)\left(-\dfrac{27}{49}\right)+3\left(-\dfrac{1}{7}\right)\)

\(=-\dfrac{8}{343}-\dfrac{162}{343}-\dfrac{147}{343}=-\dfrac{317}{343}.\)

Hence the exact value is \(\boxed{-\dfrac{317}{343}}\).

Let the roots of \(x^{3}+2x^{2}+x+7=0\) be \(\alpha,\beta,\gamma\). By Vieta’s formulae,

\(\alpha+\beta+\gamma=-2,\quad \alpha\beta+\beta\gamma+\gamma\alpha=1,\quad \alpha\beta\gamma=-7.\)

(i) We want the cubic satisfied by the three numbers

\(\dfrac{\alpha}{\beta\gamma},\quad \dfrac{\beta}{\gamma\alpha},\quad \dfrac{\gamma}{\alpha\beta}.\)

Set

\(u=\dfrac{\alpha}{\beta\gamma}.\)

Since \(\beta\gamma=\dfrac{\alpha\beta\gamma}{\alpha}=\dfrac{-7}{\alpha}\), we have

\(u=\dfrac{\alpha}{-7/\alpha}=-\dfrac{\alpha^{2}}{7}.\)

So \(\alpha^{2}=-7u\), and similarly \(\beta^{2}=-7v\), \(\gamma^{2}=-7w\), where

\(v=\dfrac{\beta}{\gamma\alpha},\quad w=\dfrac{\gamma}{\alpha\beta}.\)

Now compute the symmetric sums of \(u,v,w\).

First,

\(u+v+w=-\dfrac{1}{7}(\alpha^{2}+\beta^{2}+\gamma^{2}).\)

Also

\(\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=(-2)^2-2(1)=2.\)

Hence

\(u+v+w=-\dfrac{2}{7}.\)

Next,

\(uv+vw+wu=\dfrac{1}{49}(\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2})\).

Now

\((\alpha\beta+\beta\gamma+\gamma\alpha)^2=\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+2\alpha\beta\gamma(\alpha+\beta+\gamma),\)

so

\(1=\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+2(-7)(-2)\).

Therefore

\(\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}=1-28=-27,\)

and

\(uv+vw+wu=-\dfrac{27}{49}.\)

Finally,

\(uvw=\dfrac{\alpha\beta\gamma}{\alpha^{2}\beta^{2}\gamma^{2}}=\dfrac{1}{\alpha\beta\gamma}=-\dfrac{1}{7}.\)

Therefore the monic cubic with roots \(u,v,w\) is

\(t^{3}-(u+v+w)t^{2}+(uv+vw+wu)t-uvw=0,\)

that is

\(t^{3}+\dfrac{2}{7}t^{2}-\dfrac{27}{49}t+\dfrac{1}{7}=0.\)

Multiplying by 49 gives

\(49t^{3}+14t^{2}-27t+7=0,\)

so the stated numbers are indeed roots.

(ii) Let

\(u=\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}},\quad v=\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}},\quad w=\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}.\)

Then \(u,v,w\) are the roots from part (i) squared, so if we denote those roots by \(r_1,r_2,r_3\), then \(u=r_1^2\), \(v=r_2^2\), \(w=r_3^2\).

We use the identity

\(u+v+w=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).\)

From part (i),

\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49}.\)

Hence

\(u+v+w=\left(-\dfrac{2}{7}\right)^2-2\left(-\dfrac{27}{49}\right)=\dfrac{4}{49}+\dfrac{54}{49}=\dfrac{58}{49}.\)

So

\(\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}}+\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}}+\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}=\boxed{\dfrac{58}{49}}.\)

(iii) We want

\(\dfrac{\alpha^{3}}{\beta^{3}\gamma^{3}}+\dfrac{\beta^{3}}{\gamma^{3}\alpha^{3}}+\dfrac{\gamma^{3}}{\alpha^{3}\beta^{3}}.\)

Using the same roots \(r_1,r_2,r_3\) from part (i), this is

\(r_1^{3}+r_2^{3}+r_3^{3}.\)

Now apply

\(r_1^{3}+r_2^{3}+r_3^{3}=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)

Substituting the sums from part (i),

\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49},\quad r_1r_2r_3=-\dfrac{1}{7},\)

gives

\(r_1^{3}+r_2^{3}+r_3^{3}=\left(-\dfrac{2}{7}\right)^3-3\left(-\dfrac{2}{7}\right)\left(-\dfrac{27}{49}\right)+3\left(-\dfrac{1}{7}\right).\)

So

\(r_1^{3}+r_2^{3}+r_3^{3}=-\dfrac{8}{343}-\dfrac{162}{343}-\dfrac{147}{343}=-\dfrac{317}{343}.\)

Therefore the exact value is \(\boxed{-\dfrac{317}{343}}\).

Answer: (i) The roots are \(3,\,6,\,12\).

(ii) \(k=126\).

Let the three real roots be \(a, ar, ar^{-1}\). Then, by Vieta’s relations for

\(x^{3}-21x^{2}+kx-216=0\),

we have

\(a+ar+ar^{-1}=21\),

\(a\cdot ar+ a\cdot ar^{-1}+ ar\cdot ar^{-1}=k\),

and

\(a\cdot ar\cdot ar^{-1}=216\).

From the product,

\(a^{3}=216\), so \(a=6\).

Then the sum of the roots gives

\(6(1+r+r^{-1})=21\), so \(1+r+r^{-1}=\frac{7}{2}\).

Hence

\(r+r^{-1}=\frac{5}{2}\).

Multiplying by \(r\),

\(r^{2}-\frac{5}{2}r+1=0\),

so

\(2r^{2}-5r+2=0\).

Solving,

\(r=2\) or \(r=\frac12\). These give the same set of roots, namely

\(6,\,12,\,3\).

So the numerical values of the roots are \(3,6,12\).

Now use the sum of pairwise products to find \(k\):

\(k=6\cdot 6r + 6\cdot 6r^{-1} + (6r)(6r^{-1})\)

\(=36(r+r^{-1})+36\)

\(=36\cdot \frac{5}{2}+36=90+36=126\).

Therefore \(k=126\).

Answer: (i) With \(y=3x-1\), \(x=\dfrac{y+1}{3}\). Substitution gives \(y^3-2y-7=0\), so \(3\alpha-1\), \(3\beta-1\), \(3\gamma-1\) are its roots.

(ii) \(S_3=21\).

(iii) \(S_{-2}=\dfrac{4}{49}\).

Put \(y=3x-1\), so \(x=\dfrac{y+1}{3}\). Substituting into

\(9x^3-9x^2+x-2=0\)

gives

\(9\left(\dfrac{y+1}{3}\right)^3-9\left(\dfrac{y+1}{3}\right)^2+\dfrac{y+1}{3}-2=0.\)

Multiply by \(3\):

\((y+1)^3-3(y+1)^2+(y+1)-6=0.\)

Expanding,

\(y^3+3y^2+3y+1-3y^2-6y-3+y+1-6=0,\)

so

\(y^3-2y-7=0.\)

Thus \(3\alpha-1\), \(3\beta-1\), and \(3\gamma-1\) are the roots of this cubic.

Let these roots be \(r_1,r_2,r_3\). For \(y^3-2y-7=0\), Vieta's formulae give

\(r_1+r_2+r_3=0,\quad r_1r_2+r_2r_3+r_3r_1=-2,\quad r_1r_2r_3=7.\)

For \(S_3\), use

\(r_1^3+r_2^3+r_3^3=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)

Since \(r_1+r_2+r_3=0\),

\(S_3=3r_1r_2r_3=21.\)

For \(S_{-2}\), first find

\(S_{-1}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}=\dfrac{r_1r_2+r_2r_3+r_3r_1}{r_1r_2r_3}=-\dfrac27.\)

Also

\(S_{-1}^2=S_{-2}+2\left(\dfrac{1}{r_1r_2}+\dfrac{1}{r_2r_3}+\dfrac{1}{r_3r_1}\right).\)

Now

\(\dfrac{1}{r_1r_2}+\dfrac{1}{r_2r_3}+\dfrac{1}{r_3r_1}=\dfrac{r_1+r_2+r_3}{r_1r_2r_3}=0.\)

Therefore

\(S_{-2}=S_{-1}^2=\left(-\dfrac27\right)^2=\dfrac4{49}.\)

Answer: (a) \(b=-2,\quad c=\dfrac12\)

(b) \(d=\dfrac7{16}\)

(a)

For \(x^3+bx^2+cx+d=0\),

\(b=-(\alpha+\beta+\gamma)=-2.\)

Also

\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma).\)

Since \(\alpha\beta+\alpha\gamma+\beta\gamma=c\),

\(3=2^2-2c,\)

so

\(c=\frac12.\)

(b)

Let \(S_r=\alpha^r+\beta^r+\gamma^r\). For the cubic, Newton's sums give

\(S_3+bS_2+cS_1+3d=0.\)

Using \(b=-2\), \(c=\frac12\), \(S_1=2\), and \(S_2=3\),

\(S_3-6+1+3d=0,\)

so

\(S_3=5-3d.\)

Next,

\(S_4+bS_3+cS_2+dS_1=0.\)

Substitute \(S_4=5\):

\(5-2(5-3d)+\frac12(3)+2d=0.\)

Thus

\(-\frac72+8d=0,\)

so

\(d=\frac7{16}.\)

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