If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,

If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.

By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also $a^3+b^3=(a+b)^3-3ab(a+b)$. But both ways aren't working.
Please explain how do I solve these types of questions analytically. 
 A: Given  $$x^3+y^3=72$$
$$xy=8$$
Notice, 
$$(x+y)^3=x^3+y^3+3xy(x+y)$$ $$(x+y)^3=72+3(8)(x+y)$$
$$(x+y)^3-24(x+y)-72=0$$
Above is the cubic equation in terms of $(x+y)$ which has one real root $6$ Hence, we get $$x+y=6$$
Now, $$(x-y)^2=(x+y)^2-4xy$$
$$(x-y)=\pm\sqrt{(x+y)^2-4xy}$$
$$x-y=\pm\sqrt{(6)^2-4\times8 }$$
$$=\pm\sqrt{4}=\pm 2$$
Hence, we have 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x-y=\pm 2}}$$
A: You may solve this by brute force. Write $x=8/y$. The first equation now reads
$x^3+\frac{512}{x^3}=72$.
Multiply both sides by $x^3$, and you have
$x^6+512-72x^3=0$. 
Define $u\equiv x^3$, and the equation becomes
$u^2-72u+512=0$.
Use the quadratic formula to solve for $u$, solve back for x, and solve for y.
A: It's two equations, with two variables. We can do what we always do. You can write $y= 8/x$, and substitute into the first equation, obtaining $x^3 + 8^3/x^3 =72.$
From here, we want to write this as a polynomial, so we multiply through $x^3$ so there are no negative powers, and then move everything to the left. This gives $x^6-72x^3+512.$ This is a pretty large polynomial, but since all the exponents are multiples of 3, all is not lost. We'll set $u=x^3$ and obtain $u^2-72u + 512$. The quadratic formula gives the factorization $(u-8)(u-64)=0,$ which of course implies that $u=8$ or that $u=64$. 
Now we just go through the substitution in $u$. We write $x^3 = 8$ or $x^3 = 64$. There is only one cube root when dealing with real numbers, so we have $x=2$ or $x=4$. If $x=2$, then we have $y=8/2=4$. If $x=4$, then we have $y=8/4=2$.
So there are two solutions then, depending on the case. Either $x-y = 2$ or $x-y=-2$ depending on which pair of solutions you work with.
A: Hint: 
$$x^3+y^3=72\text{ and }x^3y^3=512.$$
You know the sum, and you know the product of two cubes.

The cubes are $\dfrac{72\pm\sqrt{72^2-4\cdot 512}}2$, $8$ and $64$.

A: If you know the basic shapes of the curves $xy=c$ (a hyperbola centered at the origin with diagonal axes) and $x^3+y^3=c$ (for $c$ positive, something like a bell curve tilted clockwise 45 degrees) then it's clear there can be exactly two solutions, exactly one solution, or no solutions. It's not so crazy to stumble upon $(2,4)$ and $(4,2)$ as solutions, and then the visual confirms there can be no more.
If you wanted to be more formal with this approach, you could establish that


*

*only in the first quadrant are both curves present

*in the first quadrant, both curves are decreasing as $x$ increases

*in the first quadrant, one curve is concave down while the other is concave up


This is enough to establish that there could be only $0$, $1$, or $2$ solutions. Since $(2,4)$ and $(4,2)$ fit the bill, you're done.
