<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] Expand the first equation to get
$$
x^2y+xy^2+x^2+2xy+y^2+x+y = 3
$$
Note that this can be rewritten
$$
x^2y+xy^2+(x+y)^2+(x+y) = 3
$$
Multiply this by $3$ and add to the second equation to get
$$
(x+y)^3+3(x+y)^2+3(x+y) = \frac{117}{8}
$$
Add $1$ to both sides to obtain
$$
(x+y+1)^3 = \frac{125}{8} = \left(\frac{5}{2}\right)^3
$$
whereupon
$$
x+y = \frac{3}{2}
$$
This means that
$$
xy+x+y+1 = (x+1)(y+1) = 2
$$
which finally yields
$$
xy = -\frac{1}{2}
$$
</s>