<s>[INST] $(x+y)(x+1)(y+1) = 3$ and $x^3 + y^3 = \frac{45}{8}$ I've came across this problem : 

If $x$ and $y$ are real numbers such that 
  $(x+y)(x+1)(y+1) = 3$ and
  $x^3 + y^3 = \frac{45}{8}$,
  find $xy$.

This is what I've tried so far:
$$x^3 + y^3 = (x+y)(x^2-xy+y^2) =\frac{45}{8}$$
So $$\frac{45}{8(x^2 - xy + y^2)} = \frac{3}{(x+1)(y+1)}$$
$$24x^2 - 24xy + 24y^2 = 45xy + 45(x+y+1)$$
$$24x^2 + 24y^2 - 45(x+y+1) = 69xy$$
But this doesn't seem to help. I tend to think it is just a matter of factorization and/or substitution, but I can't get it right.
A piece of advice would be apreciated.
Thanks!
 [/INST] See, the whole thing is symmetric, so it would be natural to switch to the symmetric polynomials. Say, $xy=a$ and $x+y=b$. Then $(x+1)(y+1)=xy+x+y+1=a+b+1$, and $x^3+y^3=(x+y)^3-3x^2y-3xy^2=b^3-3ab$.
So $b(a+b+1)=3$ and $b^3-3ab={45\over8}$. From the first equation we have $a={3\over b}-b-1$. Plugging this into the second equation, we get $b^3-3({3\over b}-b-1)b=b^3+3b^2+3b-9=(b+1)^3-10={45\over8}$. Well, the rest is kinda obvious.
</s>