## College Algebra (6th Edition)

The solutions are $x=2$ and $x=1$
$(x^{2}-3x+3)^{3/2}-1=0$ Take $1$ to the right side of the equation: $(x^{2}-3x+3)^{3/2}=1$ Raise both sides to $\dfrac{2}{3}$: $[(x^{2}-3x+3)^{3/2}]^{2/3}=1^{2/3}$ $x^{2}-3x+3=1$ Take $1$ to the left side and simplify: $x^{2}-3x+3-1=0$ $x^{2}-3x+2=0$ Solve by factoring: $(x-2)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x-1=0$ $x=1$ The solutions are $x=2$ and $x=1$