<s>[INST] Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :

In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $

MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1=x^4-x^2+1$$ $$(a+b)x^3+(ab+2)x^2+(a+b)x=-x^2$$ In order to satisfy this , $a+b=0$ and $ab+2=-1$
Hence , $$ab=-3$$

but what'll be the value of $a$ & $ b$ , which'll satisfy these conditions $?$

 [/INST] So we have $a+b=0$ and $ab=-3$
$$ a+b=0 => a=-b $$
Put this value of $a$ in the second equation
$$ -b  \cdot b=-3  $$
$$ b^2=3 => b=\pm\sqrt{3} $$
Can you now find $a$ with help of first equation?
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