Find $a, b, c$ such that element $x=a\alpha^2+b\alpha+c \in \mathbb{Q}[x]/\langle x^3+x+11 \rangle$ I'm trying to generate a fundamental unit of the number field $K=\mathbb{Q}(\alpha)$, where  $\alpha^3+\alpha+11$.
I found a non-trivial unit and I need to find $a,b,c\in\Bbb{Q}$ such that 
$$\frac{(-5\alpha^2-4\alpha+8)(6\alpha^2+6)(10\alpha^2+10\alpha)}{(8\alpha^2+8\alpha)^6(9\alpha^2+9\alpha)^{11}(10\alpha^2)}= a\alpha^2+b\alpha+c.$$
Does anyone know how to compute $a, b, c$ on Wolfram alpha or Sagemath?

A quick google search leads to the "Number field element" page of the Sage documentation, which shows that the following code does the trick in SageMath:
K.<a> = NumberField(x^3 + x + 11)
f = a.coordinates_in_terms_of_powers()
f((-5*a^2-4*a+8)*(6*a^2+6)*(8*a^2+8)^(-6)*(9*a^2+9*a)^(-11)*(10*a^2)^(-1)*(10*a^2+10*a))

