Algebra Examples

$\frac{{(a^{2} + 3)}^{\frac{1}{3}}}{{(a^{2} - 3)}^{\frac{- 1}{3}}}$

Step 1

Move ${(a^{2} - 3)}^{\frac{- 1}{3}}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

${(a^{2} + 3)}^{\frac{1}{3}} {(a^{2} - 3)}^{- \frac{- 1}{3}}$

Step 2

Move the negative in front of the fraction.

${(a^{2} + 3)}^{\frac{1}{3}} {(a^{2} - 3)}^{- - \frac{1}{3}}$

Step 3

Multiply $- 1$ by $- 1$ .

${(a^{2} + 3)}^{\frac{1}{3}} {(a^{2} - 3)}^{1 (\frac{1}{3})}$

Step 4

Multiply $\frac{1}{3}$ by $1$ .

${(a^{2} + 3)}^{\frac{1}{3}} {(a^{2} - 3)}^{\frac{1}{3}}$

Step 5

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\sqrt[3]{{(a^{2} + 3)}^{1}} {(a^{2} - 3)}^{\frac{1}{3}}$

Step 6

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\sqrt[3]{{(a^{2} + 3)}^{1}} \sqrt[3]{{(a^{2} - 3)}^{1}}$

Step 7

Anything raised to $1$ is the base itself.

$\sqrt[3]{a^{2} + 3} \sqrt[3]{{(a^{2} - 3)}^{1}}$

Step 8

Anything raised to $1$ is the base itself.

$\sqrt[3]{a^{2} + 3} \sqrt[3]{a^{2} - 3}$