Rozwiązanie

$x^8+x^2=2x^4+2x-2$
$x^8-2x^4+x^2-2x+2=0$
$\left(x^8-2x^4+1\right)+\left(x^2-2x+1\right)=0$
${\left(x^4-1\right)}^2+{\left(x-1\right)}^2=0$
${\left[\left(x^2-1\right)\left(x^2+1\right)\right]}^2+{\left(x-1\right)}^2=0$
${\left[\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\right]}^2+{\left(x-1\right)}^2=0$
${\left(x-1\right)}^2\left[{\left(x+1\right)}^2{\left(x^2+1\right)}^2+1\right]=0$
$\underset{{}_{x=1}}{{\left(x-1\right)}^2=0}\vee \underset{{}_{\text{roˊwnanie sprzeczne}}}{\underset{\ge 0}{\underbrace{{\left(x+1\right)}^2{\left(x^2+1\right)}^2}}+1=0}$

Jedynym rzeczywistym rozwiązaniem tego równania jest $x=1$.                                                $■$