equation exercise asked by Wisdom academy solved by two methods

Exercise:

Solve in $\mathbb{R}$ , $\sqrt[3]{6+\sqrt{x}}+\sqrt[3]{6-\sqrt{x}}=\sqrt[3]{3}$

Solution: Let $a=\sqrt[3]{6+\sqrt{x}}\,\,\And \,\,b=\sqrt[3]{6-\sqrt{x}}\Leftrightarrow {{a}^{3}}+{{b}^{3}}=6+\sqrt{x}+6-\sqrt{x}=12$ ..(1)

Now ${{\left( a+b \right)}^{3}}={{\left( \sqrt[3]{6+\sqrt{x}}+\sqrt[3]{6-\sqrt{x}} \right)}^{3}}={{\left( \sqrt[3]{3} \right)}^{3}}=3$ …(2)

but${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$

$\Leftrightarrow 3=12+3ab\left( a+b \right)\Leftrightarrow -9=3ab\left( a+b \right)\Leftrightarrow -3=ab\left( a+b \right)\Leftrightarrow -27=\left( 36-x \right)\left( 3 \right)\Leftrightarrow x=45$

Or : we can solve it as follows using this nice Property :

$if\ \ a+b+c=0\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

So $6+\sqrt{x}+6-\sqrt{x}-3=3\sqrt[3]{-3\left( 36-x \right)}\Leftrightarrow 3=\sqrt[3]{-108+3x}\Leftrightarrow 27=-108+3x\Leftrightarrow x=45$