Equalities with 2 given substitions Let $\alpha$ and $\beta$ be real numbers such that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ Which of the following equalities hold?

A. $\alpha^2 + \beta^2 = 2$


B. $\alpha^3 + \beta^3 = 3$


C. $\alpha^4 + \beta^4 = 6$


D. $\alpha^5 + \beta^5 = 12$


E. $\alpha^6 + \beta^6 = 18$

My attempts of solving this problem:
A.$$ \alpha^2 + \beta^2 = 2 $$

$$\alpha^2 + (-2\ \alpha \cdot \beta) + \beta^2 + 2\ \alpha\cdot\beta$$
$$=\left(\alpha + \beta \right)^2 + \left( -2\ \alpha\beta \right)  $$
We already know that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ which implies the following equation :
$1^2 + 2 \cdot 1= 3$


So we can conclude A isn't the answer

B.

$$\alpha^3 + \beta^3 = \left(\alpha +\beta \right)\left(\alpha^2 - \alpha\beta + \beta^2 \right)$$
$$1\cdot \left(3 + 1)\right) =4 $$

So B is neither the right answer.
For the following multiple choices I can't get to the final answer. Could someone help me with this?
Kind regards in advance.

Choice E. Use $\alpha^6+\beta^6 = (\alpha^3+\beta^3)^2 -2(\alpha\beta)^3 = 4^2 - 2(-1) = 18$.
