Solving two equations for $(a-b)$

Question

Let $a$ and $b$ be real numbers such that $a^{1/3}-b^{1/3}=12$ and $216ab=(a+b+8)^3$. Find the value of $a-b$.

My attempt: Rationalizing the first equation, $a-b=12(a^{2/3}+b^{2/3}+(ab)^{1/3})$. Using the second equation, $a-b=2(6a^{2/3}+6b^{2/3}+a+b+8)=2((a^{1/3}+2)^3+(b^{1/3}+2)^3-4(3a^{1/3}+3b^{1/3}+2))=2(a^{1/3}+b^{1/3}+4)([(a^{1/3}+b^{1/3}+2)^2-(2a^{1/3}+2b^{1/3}+3)])-4(3a^{1/3}+3b^{1/3}+2)$

And it keeps getting more complicated as I tried finding the value of $a^{1/3}+b^{1/3}$ using the given expression, I get $a^{1/3}+b^{1/3}=448+2a+2b$ and substituting this in does not help regenerate any term I know the value of. Any ideas as to how to solve this problem?

Answer 1

Let $x=a^{1/3}$ and $y=b^{1/3}$. Then $$x-y=12,$$ $$(6xy)^3=(x^3+y^3+8)^3.$$ The last equation is equivalent to $$x^3+y^3+8=6xy.$$

(Notice that $x$ and $y$ can’t be both non-negative, or else $x^3+y^3+8\ge 6xy$ due to AM-GM. So $x=y=2$. And $x-y=0$ instead of $12$.)

The equation $x^3+y^3+8=6xy$ is equivalent to $$(x+y+2)(x^2-xy-2x+y^2-2y+4)=0.$$

If $x+y+2=0$ then $x=5, y=-7$, and $a-b=x^3-y^3=125+343=468.$

If $x^2-xy-2x+y^2-2y+4=0$ then putting $x=y+12$ we get $$y^2+8y+124=0,$$ which has no real solutions.

Answer 2

Let $a=A^3$ and $b=B^3$ then we have

$$a^{1/3}-b^{1/3}=12 \iff A-B=12$$

and

$$216ab=(a+b+8)^3 \iff 6AB=A^3+B^3+8$$

from which we can obtain a cubic equation for $A$ or $B$ which leads to the real solution $A=5$ and $B=-7$ that is $a-b=5^3+7^3=468$.