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If $a$ and $b$ are distinct zeroes of the polynomial $x^3-2x+c$ and
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+4ab+2b^2)=y$$
Evaluate $y$

I tried for many hours but couldn't solve this question. This question is far tougher than it looks!
I made another equation using the fact that $a$ and $b$ are the distinct roots of the given polynomial. I got $a^2+ab+b^2=2$. I tried solving it as a quadratic in $a$ and then putting it in the other given equation, but the expression just got very ugly :(
Please Help!
Thanks!

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Since $a$ and $b$ are the solutions of $x^3-2x+c=0$ we have,

$a^3-2a+c=0$

&

$b^3-2b+c=0$

Subtracting the above two equations, we get,

$a^3-b^3-2(a-b)=0$

$\implies (a-b)(a^2+ab+b^2-2)=0$

$\implies a^2+ab+b^2=2$ (Since $a$ and $b$ are distinct)

Now,

$$a^2+ab+b^2=2$$
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+4ab+2b^2)=y$$

Squaring the first equation, we get,

$a^4+b^4+3a^2b^2+2a^3b+2ab^3=4$

$\implies 2a^4+2b^4+6a^2b^2+4a^3b+4ab^3=8$

Expanding and adding the other two equations, we get,

$y+3=2a^4+2b^4+6a^2b^2+4a^3b+4ab^3$

$\implies y+3=8$

$\implies \boxed{y=5}$

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  • $\begingroup$ Why $a^2+ab+b^2=2$? $\endgroup$ – AsdrubalBeltran Dec 13 '14 at 20:40
  • $\begingroup$ @AsdrubalBeltran I've added the details, you may refer to them. $\endgroup$ – MathGod Dec 13 '14 at 20:47
  • $\begingroup$ Thanks I gave you a vote + $\endgroup$ – AsdrubalBeltran Dec 13 '14 at 20:52

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