Answer: The equation with roots \(\alpha^3,\beta^3,\gamma^3\) is

\(y^3-3y^2+4y-1=0.\)

Hence

\(\alpha^6+\beta^6+\gamma^6=1.\)

Let

\(y=x^3.\)

The original cubic relation allows the equation in \(y\) to be written as

\(y^3-3y^2+4y-1=0.\)

Thus the roots of this new cubic are \(\alpha^3,\beta^3,\gamma^3\).

Let

\(p=\alpha^3,\qquad q=\beta^3,\qquad r=\gamma^3.\)

From

\(y^3-3y^2+4y-1=0,\)

Vieta's formulae give

\(p+q+r=3,\qquad pq+qr+rp=4.\)

Therefore

\(\alpha^6+\beta^6+\gamma^6=p^2+q^2+r^2.\)

Using

\(p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp),\)

we get

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