$(x+y)(x+1)(y+1) = 3$ and $x^3 + y^3 = \frac{45}{8}$

Question

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!

Answer 1

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.

Answer 2

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

$$ xy = -\frac{1}{2} $$