How do you simplify $\sqrt[3]{32 x^{12} y z^{6}}$ $\root [ 3] { 32x ^ { 12} y z ^ { 6} }$ ?

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How do you simplify $\sqrt[3]{32 x^{12} y z^{6}}$ $\root [ 3] { 32x ^ { 12} y z ^ { 6} }$ ?

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Answer 1 · 2017-05-31T03:10:33

$2 \sqrt[3]{4} \cdot x^{4} \cdot \sqrt[3]{y} \cdot z^{2}$ $2 root [3] (4) * x^4 * root[3] (y) * z^2$

Explanation:

$\sqrt[3]{32 x^{12} y z^{6}}$ $root[3] (32x^12yz^6)$
$= {(2^{5} \cdot x^{12} \cdot y \cdot z^{6})}^{\frac{1}{3}}$ $=(2^5*x^12*y*z^6)^(1/3)$
$= 2^{\frac{5}{3}} \cdot x^{\frac{12}{3}} \cdot y^{\frac{1}{3}} \cdot z^{\frac{6}{3}}$ $=2^(5/3)*x^(12/3)*y^(1/3)*z^(6/3)$
$= 2 \cdot 2^{\frac{2}{3}} \cdot x^{4} \cdot y^{\frac{1}{3}} \cdot z^{2}$ $=2*2^(2/3)*x^4*y^(1/3)*z^2$

$= 2 \sqrt[3]{4} \cdot x^{4} \cdot \sqrt[3]{y} \cdot z^{2}$ $=2 root [3] (4) * x^4 * root[3] (y) * z^2$

$= 2^{\frac{5}{3}} x^{4} y^{\frac{1}{3}} z^{2}$ $= 2^(5/3)x^4y^(1/3)z^2$

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How do you simplify $\sqrt[3]{32 x^{12} y z^{6}}$ $\root [ 3] { 32x ^ { 12} y z ^ { 6} }$ ?

1 Answer

Explanation:

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How do you simplify 3√32x12yz6\root [ 3] { 32x ^ { 12} y z ^ { 6} }?

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How do you simplify $\sqrt[3]{32 x^{12} y z^{6}}$ $\root [ 3] { 32x ^ { 12} y z ^ { 6} }$ ?