Question

which expression is equivalent to $-32^{\frac{3}{5}}$?
options: $-8$, $\frac{1}{8}$, $-\sqrt3{32^5}$, $\frac{1}{\sqrt3{32^5}}$

Explanation:

Step1: Recall the exponent rule for radicals

The general rule for a fractional exponent \(a^{\frac{m}{n}}\) is \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\) (when \(a\geq0\) for even roots). For negative bases, we consider the sign separately. Here, we have \(-32^{\frac{3}{5}}\). First, handle the base \(32\) and the exponent \(\frac{3}{5}\).

Step2: Apply the exponent rule

We know that \(32 = 2^{5}\), so let's rewrite \(32^{\frac{3}{5}}\) using the rule \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\). For \(a = 32\), \(m = 3\), \(n=5\), we have \(32^{\frac{3}{5}}=\sqrt[5]{32^{3}}\). But also, since \(32 = 2^{5}\), \(\sqrt[5]{32}=\sqrt[5]{2^{5}} = 2\), so \(32^{\frac{3}{5}}=(2^{5})^{\frac{3}{5}}=2^{3}=8\) (using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\)). Now, the original expression is \(-32^{\frac{3}{5}}\), so that's \(-8\). Let's check the options:

• Option 1: \(\frac{1}{8}\) is positive and not equal to \(-8\), so eliminate.
• Option 2: \(-8\) matches our calculation.
• Option 3: \(-\sqrt[3]{32^{5}}\) is a cube root of \(32^{5}\) with a negative sign, which is not equal to \(-8\) (since \(\sqrt[3]{32^{5}}\) is not \(8\)).
• Option 4: \(\frac{1}{\sqrt[3]{32^{5}}}\) is a reciprocal of a cube root, which is positive and not equal to \(-8\).

Answer:

\(-8\) (the option with \(-8\))