Solve for a a=( cube root of -17.1)^3

$a = {(\sqrt[3]{- 17.1})}^{3}$

Remove parentheses.

$a = {\sqrt[3]{- 17.1}}^{3}$

Simplify ${\sqrt[3]{- 17.1}}^{3}$ .

Rewrite $- 17.1$ as ${(- 1)}^{3} \cdot 17.1$ .

Rewrite.

$a = {\sqrt[3]{- 1 \cdot 17.1}}^{3}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$a = {\sqrt[3]{{(- 1)}^{3} \cdot 17.1}}^{3}$

Pull terms out from under the radical.

$a = {(- 1 \sqrt[3]{17.1})}^{3}$

Simplify the expression.

Rewrite $- 1 \sqrt[3]{17.1}$ as $- \sqrt[3]{17.1}$ .

$a = {(- \sqrt[3]{17.1})}^{3}$

Apply the product rule to $- \sqrt[3]{17.1}$ .

$a = {(- 1)}^{3} {\sqrt[3]{17.1}}^{3}$

Raise $- 1$ to the power of $3$ .

$a = - {\sqrt[3]{17.1}}^{3}$

Rewrite ${\sqrt[3]{17.1}}^{3}$ as $17.1$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{17.1}$ as ${17.1}^{\frac{1}{3}}$ .

$a = - {({17.1}^{\frac{1}{3}})}^{3}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a = - {17.1}^{\frac{1}{3} \cdot 3}$

Combine $\frac{1}{3}$ and $3$ .

$a = - {17.1}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$a = - {17.1}^{\frac{3}{3}}$

Rewrite the expression.

$a = - {17.1}^{1}$

Evaluate the exponent.

$a = - 1 \cdot 17.1$

Multiply $- 1$ by $17.1$ .

$a = - 17.1$