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Aleonysh [2.5K]
3 years ago
5

Consider the following scenario:

Mathematics
1 answer:
charle [14.2K]3 years ago
7 0
The successful completion of the exam review packet 
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Sloan [31]

Answer:  3x^2y\sqrt[3]{y}\\\\

Work Shown:

\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\

Explanation:

As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.

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