Finding the value of given Expression using property

Exercise:

Determine the value of $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$

Solution: Let $x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\Leftrightarrow x-\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=0$

Using the property that states if $a+b+c=0\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ ( its proof trivial )

So ${{x}^{3}}-\sqrt{5}-2+\sqrt{5}-2=3\left( x \right)\left( \sqrt[3]{-\left( \sqrt{5}-2 \right)\left( \sqrt{5}+2 \right)} \right)$

$\Leftrightarrow {{x}^{3}}-4=3x\sqrt[3]{-\left( 5-4 \right)}=3x\Leftrightarrow {{x}^{3}}+3x-4=0$

Put $p\left( x \right)={{x}^{3}}+3x-4$ so the divisors of $4$ are $\left\{ \pm 1,\pm 2,\pm 4 \right\}$ hence $p\left( 1 \right)=0$

Thus $p\left( x \right)=\left( x-1 \right)\left( a{{x}^{2}}+bx+c \right)=a{{x}^{3}}+\left( b-a \right){{x}^{2}}+\left( c-b \right)x-c={{x}^{3}}+3x-4$ 

So $a=1\,,b-a=0\,,c-b=3\,\,,\,\,-c=-4$ thus $\left\{ a,b,c \right\}=\left\{ 1,1,4 \right\}$ 

Hence $p\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+x+4 \right)$ thus $p\left( x \right)=0\Leftrightarrow x=1$ 

Therefore $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1$