Nice Exercise have 3 real roots and 6 complex roots enjoy Algebra

Exercise:

Solve the following equation ${{\left( \frac{{{x}^{3}}+x}{3} \right)}^{3}}+\frac{{{x}^{3}}+x}{3}=3x$

Solution: we have ${{\left( \frac{{{x}^{3}}+x}{3} \right)}^{3}}=\frac{1}{27}{{\left( {{x}^{3}}+x \right)}^{3}}=\frac{{{x}^{3}}{{\left( {{x}^{2}}+1 \right)}^{3}}}{27}={{\left( \frac{x}{3} \right)}^{3}}{{\left( {{x}^{2}}+1 \right)}^{3}}$

Also $\frac{{{x}^{3}}+x}{3}=\frac{x\left( {{x}^{2}}+1 \right)}{3}=\left( \frac{x}{3} \right)\left( {{x}^{2}}+1 \right)$

So ${{\left( \frac{{{x}^{3}}+x}{3} \right)}^{3}}+\frac{{{x}^{3}}+x}{3}=3x\Rightarrow {{\left( \frac{x}{3} \right)}^{3}}{{\left( {{x}^{2}}+1 \right)}^{3}}+\left( \frac{x}{3} \right)\left( {{x}^{2}}+1 \right)=3x$

$x\left( \frac{{{x}^{2}}}{27}{{\left( {{x}^{2}}+1 \right)}^{3}}+\frac{1}{3}\left( {{x}^{2}}+1 \right)-3 \right)=0$

$\Rightarrow x=0\,\,or\frac{{{x}^{2}}}{27}{{\left( {{x}^{2}}+1 \right)}^{3}}+\frac{1}{3}\left( {{x}^{2}}+1 \right)-3=0$

Put $w=\frac{{{x}^{2}}+1}{3}\Rightarrow 3w-1={{x}^{2}}$

So $\left( 3w-1 \right){{w}^{3}}+w-3=0\Rightarrow 3{{w}^{4}}-{{w}^{3}}+w-3=0$

$\Rightarrow 3\left( {{w}^{4}}-1 \right)+w\left( 1-{{w}^{2}} \right)=0$

$\Rightarrow 3\left( {{w}^{2}}-1 \right)\left( {{w}^{2}}+1 \right)-w\left( {{w}^{2}}-1 \right)=0$

$\Rightarrow \left( {{w}^{2}}-1 \right)\left( 3\left( {{w}^{2}}+1 \right)-w \right)=0\Rightarrow {{w}^{2}}=1\,\,or\,\,3\left( {{w}^{2}}+1 \right)-w=0$

$\Rightarrow w=\pm 1\,\,\,or\,\,3{{w}^{2}}+3-w=0$

If $w=1\Rightarrow {{x}^{2}}=3-1=2\Leftrightarrow x=\pm \sqrt{2}$

If $w=-1\Rightarrow {{x}^{2}}=-4\Leftrightarrow {{x}^{2}}=4{{i}^{2}}\Leftrightarrow x=\pm 2i$

Now Let’s solve $3{{w}^{2}}+3-w=0$

$\Rightarrow {{w}^{2}}-\frac{1}{3}w+1=0\Rightarrow {{w}^{2}}-\frac{2}{2}\frac{1}{3}w+1=0\Rightarrow {{w}^{2}}-2\frac{1}{6}w+\frac{1}{36}-\frac{1}{36}+1=0$

$\Rightarrow {{\left( w-\frac{1}{6} \right)}^{2}}+\frac{35}{36}=0\Rightarrow {{\left( w-\frac{1}{6} \right)}^{2}}=\frac{-35}{36}=\frac{35{{i}^{2}}}{36}$

$\Rightarrow w-\frac{1}{6}=\pm i\frac{\sqrt{35}}{6}\Rightarrow w=\frac{1\pm i\sqrt{35}}{6}$

Now Back to ${{x}^{2}}=3w-1\Rightarrow x=\pm \sqrt{3w-1}$

If $w=\frac{1+i\sqrt{35}}{6}\Rightarrow x=\pm \sqrt{\frac{1+i\sqrt{35}}{2}-1}=\pm \sqrt{\frac{-1+i\sqrt{35}}{2}}$

If $w=\frac{1-i\sqrt{35}}{6}\Rightarrow x=\pm \sqrt{\frac{1-i\sqrt{35}}{2}-1}=\pm \sqrt{\frac{-1-i\sqrt{35}}{2}}$

So the Solutions are :

$x\in \left\{ 0,-\sqrt{2},\sqrt{2},-2i,2i,-\sqrt{\frac{-1+i\sqrt{35}}{2}},\sqrt{\frac{-1+i\sqrt{35}}{2}},-\sqrt{\frac{-1-i\sqrt{35}}{2}},\sqrt{\frac{-1+i\sqrt{35}}{2}} \right\}$

So this equation 3 real roots and 6 complex roots