$-x^3$	=	$-27$	|: $\left(-1\right)$
$x^3$	=	$27$	|
$x$	=	$\sqrt[3]{27}$	= $3$

L={ $3$}

$2{\left(x-1\right)}^4$	=	$-162$	|:$2$
${\left(x-1\right)}^4$	=	$-81$	|

Diese Gleichung hat keine (reele) Lösung!

L={}

$\frac{1}{2}{\left(4+2x\right)}^3+5$	=	$1$
$\frac{1}{2}{\left(2x+4\right)}^3+5$	=	$1$	| $-5$
$\frac{1}{2}{\left(2x+4\right)}^3$	=	$-4$	|⋅$2$
${\left(2x+4\right)}^3$	=	$-8$	|
$2x+4$	=	$-\sqrt[3]{8}$	= $-2$

L={ $-3$}

D=R\{ $2$}

Wir multiplizieren den Nenner ${\left(x-2\right)}^2$ weg!

$\frac{4}{9}$	=	$\frac{1}{{\left(x-2\right)}^2}$	|⋅( ${\left(x-2\right)}^2$)
$\frac{4}{9}·{\left(x-2\right)}^2$	=	$\frac{1}{{\left(x-2\right)}^2}·{\left(x-2\right)}^2$
$\frac{4}{9}{\left(x-2\right)}^2$	=	$1$

$\frac{4}{9}{\left(x-2\right)}^2$	=	$1$	|⋅$\frac{9}{4}$
${\left(x-2\right)}^2$	=	$\frac{9}{4}$	|

1. Fall

2. Fall

(Alle Lösungen sind auch in der Definitionsmenge).

L={ $\frac{1}{2}$; $\frac{7}{2}$}