Basic Math Examples

\sqrt[3]{2} \cdot \sqrt[5]{3}

$\sqrt[3]{2} \cdot \sqrt[5]{3}$

Step 1

Rewrite the expression using the least common index of

15

$15$ .

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Step 1.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt[3]{2}

$\sqrt[3]{2}$ as

2^{\frac{1}{3}}

$2^{\frac{1}{3}}$ .

2^{\frac{1}{3}} \sqrt[5]{3}

$2^{\frac{1}{3}} \sqrt[5]{3}$

Step 1.2

Rewrite

2^{\frac{1}{3}}

$2^{\frac{1}{3}}$ as

2^{\frac{5}{15}}

$2^{\frac{5}{15}}$ .

2^{\frac{5}{15}} \sqrt[5]{3}

$2^{\frac{5}{15}} \sqrt[5]{3}$

Step 1.3

Rewrite

2^{\frac{5}{15}}

$2^{\frac{5}{15}}$ as

\sqrt[15]{2^{5}}

$\sqrt[15]{2^{5}}$ .

\sqrt[15]{2^{5}} \sqrt[5]{3}

$\sqrt[15]{2^{5}} \sqrt[5]{3}$

Step 1.4

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt[5]{3}

$\sqrt[5]{3}$ as

3^{\frac{1}{5}}

$3^{\frac{1}{5}}$ .

\sqrt[15]{2^{5}} \cdot 3^{\frac{1}{5}}

$\sqrt[15]{2^{5}} \cdot 3^{\frac{1}{5}}$

Step 1.5

Rewrite

3^{\frac{1}{5}}

$3^{\frac{1}{5}}$ as

3^{\frac{3}{15}}

$3^{\frac{3}{15}}$ .

\sqrt[15]{2^{5}} \cdot 3^{\frac{3}{15}}

$\sqrt[15]{2^{5}} \cdot 3^{\frac{3}{15}}$

Step 1.6

Rewrite

3^{\frac{3}{15}}

$3^{\frac{3}{15}}$ as

\sqrt[15]{3^{3}}

$\sqrt[15]{3^{3}}$ .

\sqrt[15]{2^{5}} \sqrt[15]{3^{3}}

$\sqrt[15]{2^{5}} \sqrt[15]{3^{3}}$

\sqrt[15]{2^{5}} \sqrt[15]{3^{3}}

$\sqrt[15]{2^{5}} \sqrt[15]{3^{3}}$

Step 2

Combine using the product rule for radicals.

\sqrt[15]{2^{5} \cdot 3^{3}}

$\sqrt[15]{2^{5} \cdot 3^{3}}$

Step 3

Raise

2

$2$ to the power of

5

$5$ .

\sqrt[15]{32 \cdot 3^{3}}

$\sqrt[15]{32 \cdot 3^{3}}$

Step 4

Raise

3

$3$ to the power of

3

$3$ .

\sqrt[15]{32 \cdot 27}

$\sqrt[15]{32 \cdot 27}$

Step 5

Multiply

32

$32$ by

27

$27$ .

\sqrt[15]{864}

$\sqrt[15]{864}$

Step 6

The result can be shown in multiple forms.

Exact Form:

\sqrt[15]{864}

$\sqrt[15]{864}$

Decimal Form:

1.56952263 \dots

$1.56952263 \dots$

[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$