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Leno4ka [110]
4 years ago
6

Simplify the radicals. Assume that all variables can be any real number.

Mathematics
1 answer:
larisa86 [58]4 years ago
8 0

\sqrt[3]{\frac{-27x^9}{64y^{12}}}

we take cube root separately

\sqrt[3]{-27} =\sqrt[3]{-3*-3*-3} =-3

We know cuberoot (x^3) is x

\sqrt[3]{x^9}=\sqrt[3]{x^3*x^3*x^3}=x*x*x= x^3

\sqrt[3]{64} =\sqrt[3]{4*4*4}= 4

\sqrt[3]{y^{12}} =\sqrt[3]{y^3*y^3*y^3*y^3}= y*y*y*y= y^4

\sqrt[3]{\frac{-27x^9}{64y^{12}}}=\frac{-3x^3}{4y^4}

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