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The expression $4x^3+21x-6$ has the same remainder when divided by $x-a$ or by $x+b$, where $a$ not equals to $b$. Find the value of $a^2+b^2-ab$. I have the answer but I don't know how to work through the question.

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    $\begingroup$ I think it should be $a$ not equals to $-b$. $\endgroup$ – S.H.W Jan 19 '17 at 14:41
  • $\begingroup$ Or precisely $|a|\ne|b|$ $\endgroup$ – Aditya Narayan Sharma Jan 19 '17 at 14:41
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Let $f(x)=4x^3+21x-6$

$\displaystyle f(a)=f(-b) \implies 4(a^3+b^3)+21(a+b)=0 \implies a^2-ab+b^2=\frac{-21}{4}$

Since $f(x)$ is a polynomial and it's remainder is $f(a)$ when divided by $x-a$

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Given same remainder, by Polynomial Remainder Theorem we have,

$\displaystyle\ \ \ \ \ \ f(a)=f(-b)$

$\displaystyle\Rightarrow 4a^3+21a-6=-4b^3-21b-6$

$\displaystyle\Rightarrow 4(a^3+b^3)+21(a+b)=0$

$\displaystyle\Rightarrow \frac{(a^3+b^3)}{(a+b)}=a^2+b^2-ab=\frac{-21}{4}$

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    $\begingroup$ We should assume that $a+b \not=0$ $\endgroup$ – S.H.W Jan 19 '17 at 14:42

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