Nice and easy exercise asked in كوكب الرياضيات facebook group

Exercise:

Solve in $\mathbb{R}$, $\sqrt[3]{x+10}-\sqrt[3]{x}=1$

Solution: Let $a=\sqrt[3]{x+10}$ and $b=\sqrt[3]{x}$ then $a-b=1$

But ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$

\(\Rightarrow 1=x+10-x-3ab\left( a-b \right)\Leftrightarrow 9-3ab\left( a-b \right)=0\)

So that, $ab\left( a-b \right)=3\Leftrightarrow ab=3\Leftrightarrow \sqrt[3]{x\left( x+10 \right)}=3\Leftrightarrow x\left( x+10 \right)=27$

${{x}^{2}}+10x-27=0\Leftrightarrow {{\left( x+5 \right)}^{2}}=52$ hence, $x=-5\pm 2\sqrt{13}$

If $x<0$ then $\sqrt[3]{x+10}<\sqrt[3]{10}\approx 2.15$ i.e $3=\sqrt[3]{x\left( x+10 \right)}<0$ which is impossible

Thus, $x=-5+2\sqrt{13}$ is the only acceptable real root for this equation.               Q.E.D