## College Algebra (6th Edition)

The solutions are $x=-8$ and $x=27$
$x^{2/3}-x^{1/3}-6=0$ Let $u$ be equal to $x^{1/3}$ If $u=x^{1/3}$, then $u^{2}=x^{2/3}$ Rewrite the given equation using the new variable $u$: $u^{2}-u-6=0$ Solve by factoring: $(u-3)(u+2)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-3=0$ $u=3$ $u+2=0$ $u=-2$ Substitute $u$ back to $x^{1/3}$ and solve for $x$: $u=3$ $x^{1/3}=3$ $(x^{1/3})^{3}=3^{3}$ $x=27$ $u=-2$ $x^{1/3}=-2$ $(x^{1/3})^{3}=(-2)^{3}$ $x=-8$ The solutions are $x=-8$ and $x=27$