Suppose that $\alpha$ root of the equation

Question

Suppose that $\alpha$ root of this equation: $$x^4+x^2-1=0$$

Find the value of $$\alpha ^{6}+2\alpha ^{4}$$

"I want the way, not the roots of the equation."

I tried, but I couldn't find any thing.

Answer 1

Hint: You know that $\alpha^4+\alpha^2-1 = 0$. Now, multiply both sides by $\alpha^2+1$ and see what you get.

Answer 2

$\alpha^{4}+\alpha^{2}=1$, multiplying by $\alpha^{2}$, $\alpha^{6}+\alpha^{4}=\alpha^{2}$, so adding $\alpha^{4}$ to both sides, $\alpha^{6}+2\alpha^{4}=\alpha^{2}+\alpha^{4}=1$.

Answer 3

HINT :

We have $$\alpha^4+\alpha^2=1$$and $$\alpha^6+\alpha^4=\alpha^2$$

Answer 4

If you do long division $x^6+2x^4 : x^4+x^2-1$ you get a quoatient of $x^2+1$ and a reminder of $1$. Therefore

$$x^6+2x^4=(x^2+1)(x^4+x^2-1)+1$$

Replace now $x=\alpha$.