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

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