<s>[INST] Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$(m+9)-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}-(m-9)=27$$
$$-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=9$$
$$-(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=3$$
$$-(m+9)^\frac{1}{3}(m-9)^\frac{1}{3}\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)=3$$
$$-(m+9)(m-9)\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$-(m+9)(m-9)(27)=27$$
$$m^2-81=-1$$
$$m=\pm4\sqrt{5}$$
$$\lvert m\rvert=4\sqrt{5}$$
In the third line, I cubed both sides, and then in the forth line, I expanded the left side of the equation using the binomial theorem. Is there a faster or alternative way to do this type of question? If in the question there was a higher-index root (ie. instead of the cube roots there is a higher index), I don't think my method would work, because it would take too long to apply the binomial theorem. How would solve a question in this form?
 [/INST] Let $\sqrt[3]{m+9}=a$, $-3=b$ and $-\sqrt[3]{m-9}=c$.
Hence, we have
$$a+b+c=0.$$
But $$a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=$$
$$=(a+b)^3+c^3-3ab(a+b+c)=(a+b+c)((a+b)^2-(a+b)c+c^2-3ab)=$$
$$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
We see that $$ a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2),$$
which says that $a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only.
In our case it gives
$$\sqrt[3]{m+9}=-3=-\sqrt[3]{m-9},$$ which is impossible.
Thus, our equation it's
$$a^3+b^3+c^3-3abc=0$$ or
$$m+9-27-(m-9)-9\sqrt[3]{m^2-81}=0$$ or
$$m^2-81=-1,$$
which gives the answer:
$$\{4\sqrt{5},-4\sqrt{5}\}$$
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