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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 $?$

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    $\begingroup$ Please do not delete your question once you have received an answer. If you had a reason for doing so, you may state it and I can try and reason with you, but in general it's disrespectful to the person who put up the answer for you and the audience that has read and attempted the question. $\endgroup$ Jul 2 at 1:02
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    $\begingroup$ Alt. hint: $ x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)\left((x^2+1)^2-3x^2\right)=\dots $ $\endgroup$
    – dxiv
    Jul 2 at 1:59
  • $\begingroup$ It is important to respect other answers in this community. If you felt that you did a silly mistake after reading the answer no problem just don't delete your question. Remember to delete your question, if necessary, only when no answers or hint is received in comments. Also remember if you feel the answer very helpful do remember to click the grey button to make it green showing you accept it. Don't be ashamed if you felt other people realized you have done a silly mistake. We are humans we do silly mistakes. Take a example of this $\endgroup$
    – user876009
    Jul 2 at 2:17
  • $\begingroup$ Also thanks PedroTamaroff $\endgroup$
    – user876009
    Jul 2 at 2:31
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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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  • $\begingroup$ Thanks.......I didn't thought that $\endgroup$ Jul 1 at 5:32
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Since, $a+b=0$

$\implies a=-b$

And, $ab=-3$

$\implies -b^2=-3$

$\implies b=\pm\sqrt{3}$

So, $a+b=0$

$\implies a\pm\sqrt{3}=0$

$\implies a=\mp\sqrt{3}$

$\therefore (a,b)=(\pm\sqrt{3},\mp\sqrt{3})$

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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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