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root(3)(27x^(7)y^(6))

$\sqrt[3]{27 x^{7} y^{6}}$

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Evaluate rational exponents

Full solution

Q.

\sqrt[3]{27 x^{7} y^{6}}

Identify Cube Root: Identify the cube root of each term in $\sqrt[3]{27x^7y^6}$ . For $27$ , we know that $27 = 3^3$ , so $\sqrt[3]{27} = 3$ . For $x^7$ , we can write $x^7$ as $x^6 \cdot x^1$ , and since $\sqrt[3]{x^6} = x^2$ , we have one $x$ left inside the root. For $y^6$ , it's a perfect cube, so $27$ $0$ .
Simplify Terms: Combine the simplified terms outside of the cube root. $\newline$ So we have $3x^2y^2 \times \sqrt[3]{x}$ .
Combine Simplified Terms: Write the final simplified expression. $\newline$ The final answer is $3x^2y^2 \cdot \sqrt[3]{x}$ .

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