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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?

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  • $\begingroup$ Your solution is not full. You need to substitute $x=4\sqrt5$ and $x=-4\sqrt5$ in the original equation and to show that the equality is true. $\endgroup$ – Michael Rozenberg Aug 8 '17 at 19:06
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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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  • $\begingroup$ In the second line, I think it should be $(a+b)^3+c^3−3ab(a+b+c)$. Also, how did you go from $(a+b)^3+c^3−3ab(a+b+c)$ to $(a+b+c)((a+b)^2−(a+b)c+c^2−3ab)$ in the second line? $\endgroup$ – MattMath Aug 8 '17 at 20:22
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    $\begingroup$ @MattMath I used $x^3+y^3=(x+y)(x^2-xy+y^2)$. $\endgroup$ – Michael Rozenberg Aug 9 '17 at 2:30
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Hint:

Let $\sqrt[3]{m+3}=a$ and $\sqrt[3]{m-3}=b$

$a-b=?$

$a^3-b^3=?$

Can you find $ab$ and $a^3+b^3=(a+b)\{(a-b)^2+ab\}$

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Go ahead and cube: $$m+9=27+27\sqrt[3]{m-9}+9\sqrt[3]{(m-9)^2}+m-9.$$ Denote: $\sqrt[3]{m-9}=t$ to get: $$t^2+3t+1=0 \Rightarrow t=\frac{-3\pm \sqrt{5}}{2}$$ Now: $$m-9=t^3$$ will produce the answers.

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