For Q1, you have x<sup>3</sup> + x<sup>2</sup> + 2 = 0, with roots alpha, beta and gamma. You want an equation with roots alpha<sup>4</sup>, beta<sup>4</sup> and gamma<sup>4</sup>. So, let u = x<sup>4</sup>, and so x = u <sup>1/4</sup>. Substituting, we get:
u<sup>3/4</sup> + u<sup>1/2</sup> + 2 = 0
u<sup>3/4</sup> = -(2 + u<sup>1/2</sup>)
On squaring both suides, this becomes u * sqrt(u) = 4 + 4 * sqrt(u) + u
(u - 4) * sqrt(u) = u + 4
Squaring both sides again, we get u(u<sup>2</sup> - 8u + 16) = u<sup>2</sup> + 8u + 16
Hence, the equation u<sup>3</sup> - 9u<sup>2</sup> + 8u - 16 = 0 has the required roots