Answer: \(\alpha^6+\beta^6+\gamma^6+\delta^6=7.\)

Let \(y=x^3\). Since the original equation may be written so that \(x^4=x^3+1\), we have

\(x^{12}=(x^4)^3=(x^3+1)^3.\)

Thus

\(y^4=(y+1)^3.\)

Expanding and rearranging gives the quartic whose roots are \(\alpha^3,\beta^3,\gamma^3,\delta^3\):

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

Let these roots be \(y_1,y_2,y_3,y_4\). By Vieta's formulae,

\(y_1+y_2+y_3+y_4=1,\qquad \sum_{i\lt j}y_iy_j=-3.\)

But

\(\alpha^6+\beta^6+\gamma^6+\delta^6=y_1^2+y_2^2+y_3^2+y_4^2.\)

Using

\(y_1^2+y_2^2+y_3^2+y_4^2=(y_1+y_2+y_3+y_4)^2-2\sum_{i\lt j}y_iy_j,\)

we get

\(\alpha^6+\beta^6+\gamma^6+\delta^6=1^2-2(-3)=7.\)