Originally Posted by

**ebaines** You seem to have trid to say that:

$\displaystyle (\frac 7 2 x - y)^3 = (\frac 7 2 x)^3 + (-y)^3$,

but that's not right. If you multiply $\displaystyle (A+B)^3$ out you find it's not $\displaystyle A^3 = B^3$,but rather: $\displaystyle (A+B)^3 = (A+B)(A+B)(A+B) = A^3 + 3A^2B + 3AB^2 + B^3$.

I woud suggest rather than use that approach try this: from the first equation you can get y by itself:

$\displaystyle \frac {-7} 2 x - y = -18 \ \to\ y = \frac {-7} 2 x + 18$.

Now substitute this value for 'y' into the second equation:

$\displaystyle 8x^2 -2y^3 = 0 \ \to \ 8x^2 -2(\frac {-7} 2 x + 18)^3 = 0$

Can you take it from here?