1Cademy - Simplifying $$\sqrt{180m^9n^{11}}$$, $$\sqrt[3]{72x^6y^5}$$, and $$\sqrt[4]{80x^7y^4}$$

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Simplifying $\sqrt{63u^3v^5}$

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Simplifying $\sqrt{180m^9n^{11}}$ , $\sqrt[3]{72x^6y^5}$ , and $\sqrt[4]{80x^7y^4}$

To simplify the radical expressions $\sqrt{180m^9n^{11}}$ , $\sqrt[3]{72x^6y^5}$ , and $\sqrt[4]{80x^7y^4}$ , use the Product Property of Roots to factor out the largest perfect powers matching their respective indices.

For the square root $\sqrt{180m^9n^{11}}$ , the largest perfect square factor of $180$ is $36$ ( $6^2$ ), of $m^9$ is $m^8$ , and of $n^{11}$ is $n^{10}$ . Rewrite as $\sqrt{36m^8n^{10} \cdot 5mn}$ and separate the radicals. Simplify the perfect square part to obtain $6m^4|n^5|\sqrt{5mn}$ . Absolute value signs are needed around $n^5$ because the index is even and the resulting power is odd.

For the cube root $\sqrt[3]{72x^6y^5}$ , the greatest perfect cube factor of $72$ is $8$ ( $2^3$ ), of $x^6$ is $x^6$ , and of $y^5$ is $y^3$ . Rewrite as $\sqrt[3]{8x^6y^3 \cdot 9y^2}$ . Split the radicals and simplify the perfect cube to get $2x^2y\sqrt[3]{9y^2}$ . No absolute value is needed since the index is odd.

For the fourth root $\sqrt[4]{80x^7y^4}$ , the largest perfect fourth power factor of $80$ is $16$ ( $2^4$ ), of $x^7$ is $x^4$ , and of $y^4$ is $y^4$ . Rewrite as $\sqrt[4]{16x^4y^4 \cdot 5x^3}$ . Split the radicals and simplify to obtain $2|xy|\sqrt[4]{5x^3}$ . Absolute value signs are required around both variables because the index is even and their extracted powers are odd.

Updated 2026-05-25

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OpenStax Intermediate Algebra

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