(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substituting \(y = (1-y)^4\), we expand using the binomial theorem: \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearranging gives \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The sum \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Using the formula for the sum of squares, \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).

(a) We use the identity \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Substituting the given values, \(3 = 2^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Solving, \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{1}{2}\).

(b) We use the identity \(\alpha^2(\beta + \gamma + \delta) + \beta^2(\alpha + \gamma + \delta) + \gamma^2(\alpha + \beta + \delta) + \delta^2(\alpha + \beta + \gamma)\).

This simplifies to \(\alpha^2(2-\alpha) + \beta^2(2-\beta) + \gamma^2(2-\gamma) + \delta^2(2-\delta)\).

Substituting the given values, \(2(3) - 4 = 2\).

(c)(i) Using the polynomial equation, \(6S_4 - 12S_3 + 3S_2 + 2S_1 + 24 = 0\).

Substitute \(S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_4 = \frac{11}{6}\).

(c)(ii) Using the polynomial equation, \(6S_5 - 12S_4 + 3S_3 + 2S_2 + S_1 = 0\).

Substitute \(S_4 = \frac{11}{6}, S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_5 = -\frac{4}{3}\).

(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substitute \(y^{3/4} = (1-y)^4\) to get \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearrange to form \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The value of \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Use the formula for the sum of squares: \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).

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

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

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

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived equation and given \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27\):

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).

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

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

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

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived quartic equation:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).

(a) Let \(y = x^2\), so \(x = y^{\frac{1}{2}}\). Substitute into the original equation:

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

Rearrange and square both sides to eliminate the square root:

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

Expand and simplify to obtain:

\(y^4 - 2y^3 + 11y^2 - 14y + 25 = 0\).

The value of \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\) is 2.

(b) Using the relation \(\alpha^2 \beta^2 \gamma^2 \delta^2 = 25\), we have:

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

Substitute the known values to get \(\frac{14}{25}\).

(c) Using the identity:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)^2 - 2(\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \alpha^2 \delta^2 + \beta^2 \gamma^2 + \beta^2 \delta^2 + \gamma^2 \delta^2)\).

Substitute the known values to get \(2^2 - 2(11) = -18\).

(a) Let \(y = x^{-2}\), then \(x = y^{-\frac{1}{2}}\). Substitute into the original equation:

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

Multiply through by \(y^2\) to clear fractions:

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

Rearrange to form a quartic equation:

\(36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0\)

The value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\) is \(-\frac{8}{9}\).

(b) Using the relation \(\alpha \beta \gamma \delta = 6\), we find:

\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2}{\alpha^2 \beta^2 \gamma^2 \delta^2}\)

\(\beta^2 \gamma^2 \delta^2 + \alpha^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 = -32\)

(c) Using the formula for the sum of squares:

\(\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2})\)

\(= \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)\)

= \(-\frac{61}{162}\)

(a) Let \(y = x^3\). Then \(y^{4/3} - 2y^{1/3} - 1 = 0\). Substitute \(y = x^3\) into the equation to get \(y^4 - 8y^3 - 12y^2 - 6y - 1 = 0\). The value of \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3\) is 8.

(b) The expression \(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + \frac{1}{\delta^3}\) can be rewritten using the identity \(\frac{\alpha^3 \beta^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \gamma^3 + \beta^3 \gamma^3 \delta^3 + \alpha^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \delta^3}{\alpha^3 \beta^3 \gamma^3 \delta^3} = \frac{6}{-1} = -6\).

(c) Using the original equation, \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) + 4 = 2(8) + 4 = 20\).

Answer: (a) \(y^4-2y^3+4y^2+2y+11=0\)

(b) \(S_2=-4\)

(c) \(S_4=-76\)

(a)

Let \(y=2x+1\). Then \(x=\frac12(y-1)\). Substitute into the original equation:

\(\left(\frac{y-1}{2}\right)^4+\left(\frac{y-1}{2}\right)^3+\left(\frac{y-1}{2}\right)^2+\frac{y-1}{2}+1=0.\)

That is

\(\frac1{16}(y-1)^4+\frac18(y-1)^3+\frac14(y-1)^2+\frac12(y-1)+1=0.\)

Expanding and simplifying gives

\(\frac1{16}y^4-\frac18y^3+\frac14y^2+\frac18y+\frac{11}{16}=0.\)

Multiplying by \(16\),

\(y^4-2y^3+4y^2+2y+11=0.\)

(b)

For the transformed quartic \(y^4-2y^3+4y^2+2y+11=0\), the sum of roots is \(S_1=2\), and the sum of products in pairs is \(4\). Hence

\(S_2=(\text{sum of roots})^2-2(\text{sum of products in pairs})=2^2-2(4)=-4.\)

(c)

Each transformed root \(y\) satisfies

\(y^4-2y^3+4y^2+2y+11=0.\)

Sum this equation over the four roots:

\(S_4-2S_3+4S_2+2S_1+44=0.\)

Using \(S_1=2\), \(S_2=-4\), and \(S_3=-22\),

\(S_4-2(-22)+4(-4)+2(2)+44=0,\)

so \(S_4+76=0\), and therefore

\(S_4=-76.\)

Mark scheme