$x + y = 7 $and $x^3 - y^3 = 37$, find $xy$

Question

$x+y=7$
$x^3-y^3 = 37$

Find $xy$

I have tackled this one but I am stuck, when I get to this point: $$(x-y)((x+2)^2 -xy) = 37$$ I would be glad if you could give me any suggestions.

Answer 1

Setting $y=7-x$, we obtain $$ 0=x^3-y^3-37=(2x^2 - 13x + 95)(x - 4). $$ For integer solutions, the quadratic term has no integer root, so that $x=4$ and $y=3$, hence $xy=12$. Over the real numbers, this is also the unique solution, since $2x^2-13x+95=0$ has two complex, non-real solutions.

Answer 2

$x-y=\pm \sqrt{(x+y)^2-4xy}=\pm \sqrt{49-4xy}$

$x^3-y^3=(x-y)(x^2+y^2+xy)=(x-y)\{(x+y)^2-xy\}$

$\implies \pm \sqrt{49-4xy}(49-xy)=37 $

Hope you can solve this equation to get $xy=12$ as you wanted to solve for $xy$ instead of $x$ and $y$ seperately. Still, I would go with @Dietrich Burde 's solution as you have not stated whether you want natural number solutions or not.

Answer 3

$(49-4xy)(49-xy)^2=37^2$ , $37$ is prime. So $49-4xy=1$ then $$xy=12$$

Answer 4

Since $(x^3-y^3)^2$ is symmetric in $x$ and $y$, you can express it as a polynomial in terms of $x+y$ and $xy$ :

$(x^3-y^3)^2 = x^6 - 2x^3y^3 + y^6 = (x+y)^6 - (6x^5y+15x^4y^2+22x^3y^3+15x^2y^4+6xy^5) \\ = (x+y)^6 - 6xy(x+y)^4 + (9x^4y^2+14x^3y^3+9x^2y^4) \\ = (x+y)^6 - 6xy(x+y)^4 + 9(xy)^2(x+y)^2 - 4(xy)^3$

Plugging the values $x+y = 7$ and $x^3-y^3 = 37$ gives you the equation

$4(xy)^3 - 441(xy)^2 +14406(xy)-116280 = 0$

The only rational root of this cubic polynomial is $12$, so this factors as

$(xy-12)(4(xy)^2-393(xy)+9690) = 0$

And now you can check that the second factor doesn't have real roots.

Finally, getting a value for $xy$ (here, $12$) determines the values of $x$ and $y$ completely, because

$x = (x+y)/2 + (x^3-y^3)/(2((x+y)^2-xy))= 7/2 + 37/(2.(7^2-12)) = 4$ and
then $y = (x+y)-x = 7-4 = 3$