Q.If $\alpha, \beta$ are the roots of the equation $x^2 + x + 1 = 0$, then prove that $\alpha^4 + \beta^4 + \alpha^{-1}\beta^{-1} = 0$.

Get $\alpha+\beta$ and $\alpha\beta$ from the coefficients, build up $\alpha^2+\beta^2$ then $\alpha^4+\beta^4$, and evaluate $\alpha^{-1}\beta^{-1}=1/(\alpha\beta)$.

Since $\alpha,\beta$ are roots of $x^2+x+1=0$:

$\alpha+\beta = -1, \qquad \alpha\beta = 1$

Step 1 — find $\alpha^2+\beta^2$:

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

Step 2 — find $\alpha^4+\beta^4$:

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

Step 3 — find $\alpha^{-1}\beta^{-1}$: …