## Algebra 1: Common Core (15th Edition)

$x^{3}\cdot y^{2}\cdot x^{-3}=y^{2}$
x-terms The RHS has no $x^{n}$ terms, so we can say $x^{0}$ is a factor of the RHS. The LHS has $x^{3}\cdot x^{m}=x^{3+m}$ Since $3+m=0$, m must be $-3.$ y-terms. The exponent on the RHS is 2, so it must be 2 on the LHS $x^{3}\cdot y^{2}\cdot x^{-3}=y^{2}$