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If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $h(x)=xf(x^{3})+x^{2}g(x^{6})$ is divisible by $x^{2}+x+1$, then choose the correct option:

$A. f(1)=g(1)$

$B. f(1) $ is not equal to $g(1)$

$C. f(1)$ and $g(1)$ are non-zero, $f(1)=g(1)$

$D. f(1)=-g(1)$, $f(1)$ and $g(1)$ are non-zero.

What I did: Since $x^{2}+x+1$ divides the $h(x)$, $\omega$ and $\omega^{2}$ are roots of $h(x)$. Therefore, $\omega f(1)+\omega^{2}g(1)=0$ and $\omega^{2} f(1)+\omega g(1)=0$ which gives $f(1)=-g(1)$. But the answer is option $A$. Please help...

NOTE: $\omega$ is $e^{\iota 2\pi /3}$, $\iota =(-1)^{1/2}$

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up vote 3 down vote accepted

You're correct that $f(1)=-g(1)$. But, as Ian points out above, we also have that $g(1)=f(1)$. Put these two facts together, and you'll see that the only correct answer is A!

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Thanks... Didn't note that !! – Apurv Dec 15 '13 at 6:45

EDIT: I think you mean $\omega=(1)^{1/3}$. You're left with $$ \omega f(1)+\omega^2g(1)=\omega^2 f(1)+\omega g(1)\implies (\omega^2-\omega)(g(1)-f(1))=0 $$

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No. $(-1)^{1/3}$ doesn't satisfy $x^2+x+1=0$, it satisfies $x^2-x+1=0$. On the other hand, the manipulations you've got here are correct (and indeed are only true if $\omega = e^{2\pi i/3}$). – Mike Miller Dec 15 '13 at 6:41
No, the minimal polynomial of the third roots of unity is $x^2+x+1=0$. – Ian Coley Dec 15 '13 at 6:43 is the cube root of unity... – Apurv Dec 15 '13 at 6:43
That's correct. $(-1)^{1/3}$ is not a third root of unity, it's a sixth root of unity. – Mike Miller Dec 15 '13 at 6:44
Mike is right on the money. There is an error in one of the answers above. – Doc Dec 15 '13 at 6:45

It seems to me that $f(1)$ and $g(1)$ satisfy two equations which are linearly independent. Hence $f(1) = g(1) = 0$. That implies (A), as well as your answer.

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