<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] The equation $t^2-(a+b)t+ab=0$ can be written $(t-a)(t-b)=0$ and thus has solutions $t=a$ and $t=b$.
In your case this equation becomes $t^2-3=0$ which has solutions $t=\pm\sqrt3,$ so $a,b=\pm\sqrt3.$
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