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 $?$


Since, $a+b=0$
$\implies a=-b$
And, $ab=-3$
$\implies -b^2=-3$
$\implies b=\pm\sqrt{3}$
So, $a+b=0$
$\implies a\pm\sqrt{3}=0$
$\implies a=\mp\sqrt{3}$
$\therefore (a,b)=(\pm\sqrt{3},\mp\sqrt{3})$
