$f(x)=x^4-2x^2-8f(x)=x^4-2x^2-8$

Setze die Gleichung gleich $00$.

$0=x^4-2x^2-80=x^4-2x^2-8$

$0=(x^2{)}^2-2x^2-80=(x^2)^2-2x^2-8$

Substituiere: $z=x^{2}z=x^2$.

$0=z^2-2z-80=z^2-2z-8$

Löse die so gewonnene Gleichung mit Hilfe der Mitternachtsformel.

$z_{\mathrm{1,2}}={\displaystyle \frac{2\pm \sqrt{2^2-4\cdot 1\cdot (-8)}}{2\cdot 1}}z\_\{1\{,\}2\}=\backslash dfrac\{2\; \backslash pm\backslash sqrt\{2^2-4\backslash cdot1\backslash cdot(-8)\}\}\{2\backslash cdot1\}$

Jetzt wird noch vereinfacht.

$z_{\mathrm{1,2}}={\displaystyle \frac{2\pm \sqrt{36}}{2}}z\_\{1\{,\}2\}=\backslash dfrac\{2\; \backslash pm\backslash sqrt\{36\}\}\{2\}$

$z_1={\displaystyle \frac{2+6}{2}}=4z\_\{1\}=\backslash dfrac\{2+6\}\{2\}=4$

$z_2={\displaystyle \frac{2-6}{2}}=-2z\_\{2\}=\backslash dfrac\{2-6\}\{2\}=-2$

$f(x)={\displaystyle \frac{1}{2}}{x}^6+x^3-4f(x)=\backslash dfrac\{1\}\{2\}x^6+x^3-4$

Setzte die Gleichung gleich $00$.

$0={\displaystyle \frac{1}{2}}(x^3{)}^2+x^3-40=\backslash dfrac12(x^3)^2+x^3-4$

Substituiere: $u=x^{3}u=x^3$.

$0={\displaystyle \frac{1}{2}}{u}^2+u-40=\backslash dfrac12u^2+u-4$

Wende nun die Mitternachtsformel an um die Nullstellen der substituierten Gleichung zu berechnen.

$u_{\mathrm{1,2}}={\displaystyle \frac{-1\pm \sqrt{1^2-4\cdot \frac{1}{2}\cdot (-4)}}{2\cdot \frac{1}{2}}}u\_\{1\{,\}2\}=\backslash dfrac\{-1\backslash pm\backslash sqrt\{1^2-4\; \backslash cdot\; \backslash frac\{1\}\{2\}\; \backslash cdot(-4)\}\}\{2\; \backslash cdot\; \backslash frac\{1\}\{2\}\}$

Jetzt wird noch vereinfacht.

$u_{\mathrm{1,2}}={\displaystyle \frac{-1\pm \sqrt{1+8}}{1}}u\_\{1\{,\}2\}=\backslash dfrac\{-1\backslash pm\backslash sqrt\{1+8\}\}\{1\}$

$u_{\mathrm{1,2}}=-1\pm \sqrt{9}u\_\{1\{,\}2\}=-1\backslash pm\backslash sqrt\{9\}$

$u_{\mathrm{1,2}}=-1\pm 3u\_\{1\{,\}2\}=-1\backslash pm3$

Berechne $u_{1}u\_1$ und $u_{2}u\_2$.

$u_1=2u\_1=2$ und $u_2=-4u\_2=-4$

$u_1=2\iff x_1=\sqrt[3]{2}\approx \mathrm{1,26}u\_1=2\; \backslash Leftrightarrow\; x\_1=\backslash sqrt[3]\{2\}\backslash approx1\{,\}26$

$u_2=-4\iff x_2=\sqrt[3]{-4}\approx -\mathrm{1,59}u\_2=-4\; \backslash Leftrightarrow\; x\_2=\backslash sqrt[3]\{-4\}\backslash approx\{-1\{,\}59\}$