## Precalculus (6th Edition)

$\color{blue}{15\sqrt[3]{75}}$
Note that $25=5^2$, $(-3)^4=81$ and $81=3^4$. Thus, the given expression is equivalent to: $=\sqrt[3]{5^2(81)(5^3)} \\=\sqrt[3]{5^2\cdot 3^4\cdot 5^3}$ Use the rule $a^m\cdot a^n = a^{m+n}$ to obtain: $=\sqrt[3]{3^4\cdot5^{2+3}} \\=\sqrt[3]{3^4\cdot 5^5}$ Factor the radicand so that at least one factor is a perfect cube to obtain: $=\sqrt[3]{((3^3\cdot5^3)(3\cdot5^2)} \\=\sqrt[3]{(3^3\cdot5^3)(3\cdot25)} \\=\sqrt[3]{(3^3\cdot5^3)(75)}$ Bring out the cube root of the perfect cube factors to obtain: $=3\cdot5\sqrt[3]{75} \\=\color{blue}{15\sqrt[3]{75}}$