Solve cubic root equation

Question

$ \sqrt[3]{x+5} - \sqrt[3]{x-5} = 1 $

How many values of x satisfy the equation?

Answer 1

We know that $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ Also, $$a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2\geq0,$$ where the equality occurs for $a=b=c.$

Now, let $\sqrt[3]{5+x}=a$, $\sqrt[3]{5-x}=b$ and $-1=c$.

Thus, since $a+b+c=0$ and $a=b=c$ is impossible, we obtain that our equation is equivalent to $$5+x+5-x-1+3\sqrt[3]{(5-x)(5+x)}=0$$ or $$(5-x)(5+x)=-27,$$ which gives the set of roots: $$\{-2\sqrt{13},2\sqrt{13}\}.$$ Id est, we have two values of $x$.

Answer 2

Well, just one and the answer is given by Mathematica!

Answer 3

Substituting $$a=\sqrt[3]{x+5}$$ and $$b=\sqrt[3]{x-5}$$ so we get $$a-b=1$$ and $$a^3-b^3=10$$ and this is easy to solve.