Solving equation for a field extension 
Let $F(x)=x^{3}+2 x-2,$ let $\alpha \in \mathbb{C}$ be a root of $F,$ and let $K=\mathbb{Q}(\alpha)$
Find $a, b, c \in \mathbb{Q}$ such that $\alpha^{4}=a \alpha^{2}+b \alpha+c$

We have that [$K:\mathbb{Q}]=3$.
I know that such a thing is possible, but do not know in what manner to proceed. Please leave a hint only as this is a homework assignment.
 A: $\alpha$ is a root of $F$, so $\alpha^3 + 2\alpha - 2 = ?$ And you should be able to solve that for $\alpha^4$ (by multiplying by one thing, then moving all non-$\alpha^4$ terms to the right-hand side).
A: Since $\alpha$ is a root of
$f(x) = x^3 + 2x - 2, \tag 1$
we have
$\alpha^3 + 2\alpha - 2 = 0, \tag 2$
or
$\alpha^3 = -2\alpha + 2; \tag 3$
thus,
$\alpha^4 = -2\alpha^2 + 2\alpha; \tag  4$
therefore,
$a = -2, \tag 5$
$b = 2, \tag 6$
and
$c = 0. \tag 7$
In order to show uniqueness, suppose there are
$q, r, s \in \Bbb Q \tag 8$
with
$\alpha^4 = q\alpha^2 + r\alpha + s; \tag 9$
we subract (4) from (9):
$0 = (q + 2)\alpha^2 + (r - 2)\alpha + s; \tag{10}$
thus $\alpha$ must satisfy the polynomial
$g(x) = (q + 2)x^2 + (r - 2)x + s \in \Bbb Q[x]; \tag{11}$
we observe that $f(x)$ is irreducible over $\Bbb Q$ via the Eisenstein criterion with prime $p = 2$; therefore it is minimal for $\alpha$ over $\Bbb Q$; but this contradicts
$g(\alpha) = 0 \tag{12}$
unless
$g(x) = 0; \tag{13}$
therefore (13) binds and thus
$q = -2 = a, \tag{14}$
$r = 2 = b, \tag{15}$
and
$s = c = 0, \tag{16}$
and we conclude the expression (4) for $\alpha^4$ is unique.
