Answer
The solutions are $x=-1$ and $x=\sqrt[3]{3}$
Work Step by Step
$x^{6}-2x^{3}-3=0$
Let $u$ be equal to $x^{3}$:
$u=x^{3}$, so $u^{2}=x^{6}$
Substitute $x^{6}$ by $u^{2}$ and $x^{3}$ by $u$ in the original equation:
$u^{2}-2u-3=0$
Solve this equation by factoring:
$(u+1)(u-3)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u+1=0$
$u=-1$
$u-3=0$
$u=3$
Substitute $u$ back to $x^{3}$ and solve for $x$:
$u=-1$
$x^{3}=-1$
$x=\sqrt[3]{-1}$
$x=-1$
$u=3$
$x^{3}=3$
$x=\sqrt[3]{3}$
The solutions are $x=-1$ and $x=\sqrt[3]{3}$
