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

Question

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.

Answer 1

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 $64$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x-y=\pm 2}}$$

Answer 5

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.