## College Algebra (11th Edition)

$x=4$ or $x=1$
We are given: $(x-2)^{2/3}=x^{1/3}$ We raise both sides to the 3rd power: $[(x-2)^{2/3}]^{3}=(x^{1/3})^{3}$ $(x-2)^{2}=x^1$ $(x-2)(x-2)=x$ And distribute: $x^{2}-4x+4=x$ $x^{2}-5x+4=0$ And factor: $(x-4)(x-1)=0$ Equate each factor to zero, then solve each equation for $x$: $(x-4)=0$ or $(x-1)=0$ $x=4$ or $x=1$