nice exercise asked in the brilliant website

Exercise:

Determine the value of $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$

Solution: Let $x=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$

Take $a+b+c=0\Leftrightarrow a+b=-c$ cubic both sides to get

${{\left( a+b \right)}^{3}}=-{{c}^{3}}\Leftrightarrow {{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}=-{{c}^{3}}\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3\left( {{a}^{2}}b+a{{b}^{2}} \right)=0$

$\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3ab\left( a+b \right)=0\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

So $x-\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{20-14\sqrt{2}}=0$

$\Leftrightarrow {{x}^{3}}-\left( 20+14\sqrt{2} \right)-\left( 20-14\sqrt{2} \right)=3x\sqrt[3]{\left( 20+14\sqrt{2} \right)\left( 20-14\sqrt{2} \right)}$

$\Leftrightarrow {{x}^{3}}-40=3x\sqrt[3]{\left( 400-392 \right)}=6x$ $\Leftrightarrow {{x}^{3}}-6x-40=0$

But the divisors of $40$ are $\left\{ \pm 1,\pm 2,\pm 4,\pm 5,\pm 8,\pm 20,\pm 40 \right\}$

Consider $p\left( x \right)={{x}^{3}}-6x-40$

$p\left( 1 \right)=1-6+40\ne 0$ ,$p\left( 2 \right)=8-12+40\ne 0$ ,$p\left( 4 \right)=64-24-40=0$

Hence $x=4$ is a root for $p\left( x \right)$ thus $p\left( x \right)=\left( x-4 \right)\left( a{{x}^{2}}+bx+c \right)$

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

So $a=1\,\,,\,b-4a=0\,\,,c-4b=-6\,\,\And \,-4c=-40$ i.e $\left( a,b,c \right)=\left\{ 1,4,10 \right\}$

Hence $p\left( x \right)=\left( x-4 \right)\left( {{x}^{2}}+4x+10 \right)$

$p\left( x \right)=0\Leftrightarrow \left( x-4 \right)\left( {{x}^{2}}+4x+10 \right)=0\Leftrightarrow x=4\,\,or\,{{x}^{2}}+4x+10=0$

Thus $x=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$