# Solving two equations for $(a-b)$

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?

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.

• Ah shoot, I didn't recognise the identity form of $a^3+b^3+c^3=3abc$. I understand now, thanks! Commented Feb 11 at 10:22

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 $$a-b=5^3+7^3=468$$.