Finding the value of $y=b^2(3a^2+4ab+2b^2)$ if $a^2(2a^2+4ab+3b^2)=3$ and $a$ and $b$ are distinct zeros of $x^3-2x+c$ 
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!

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,   
$y+3=2a^4+2b^4+6a^2b^2+4a^3b+4ab^3$  
$\implies y+3=8$  
$\implies \boxed{y=5}$
