Zdanie drugie

${\left(\sqrt{2}+x\right)}^3-3x\left(\sqrt{2}x+\frac{1}{3}{x}^2\right)-6x\stackrel{?}{=}2\sqrt{2}$

${\left(\sqrt{2}+x\right)}^3={\left(\sqrt{2}\right)}^3+3\cdot {\left(\sqrt{2}\right)}^2\cdot x+3\cdot \sqrt{2}\cdot x^2+x^3=\sqrt{8}+3\cdot 2\cdot x+3\sqrt{2}{x}^2+x^3=$

$=2\sqrt{2}+6x+3\sqrt{2}{x}^2+x^3$

${\left(\sqrt{2}+x\right)}^3-3x\left(\sqrt{2}x+\frac{1}{3}{x}^2\right)-6x=2\sqrt{2}+6x+3\sqrt{2}{x}^2+x^3-3\sqrt{2}{x}^2-1x^3-6x=2\sqrt{2}$

Zdanie jest prawdziwe.