How do you simplify $2 \sqrt[3]{16 x^{2} y^{7}}$ $2\root[ 3] { 16x ^ { 2} y ^ { 7} }$ ?

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How do you simplify $2 \sqrt[3]{16 x^{2} y^{7}}$ $2\root[ 3] { 16x ^ { 2} y ^ { 7} }$ ?

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Answer 1 · 2017-10-09T00:38:37

$= \sqrt[3]{2^{7} x^{2} y^{7}} = 2^{\frac{7}{3}} x^{\frac{2}{3}} y^{\frac{7}{3}}$ $=root(3)(2^7x^2y^7)=2^(7/3)x^(2/3)y^(7/3)$

Explanation:

$2 \cdot {(16 x^{2} y^{7})}^{\frac{1}{3}} = 2 \cdot 16^{\frac{1}{3}} \cdot x^{\frac{2}{3}} \cdot y^{\frac{7}{3}}$ $2*(16x^2y^7)^(1/3)=2*16^(1/3)*x^(2/3)*y^(7/3)$
$= 2 \cdot 2^{\frac{4}{3}} \cdot x^{\frac{2}{3}} \cdot y^{\frac{7}{3}}$ $=2*2^(4/3)*x^(2/3)*y^(7/3)$
$= 2^{\frac{7}{3}} \cdot x^{\frac{2}{3}} \cdot y^{\frac{7}{3}}$ $=2^(7/3)*x^(2/3)*y^(7/3)$
$= \sqrt[3]{2^{7} x^{2} y^{7}})$ $=root(3)(2^7x^2y^7))$

Aliter :
$= {\sqrt[3]{2}}^{3} \cdot \sqrt[3]{2^{4} \cdot x^{2} \cdot y^{7}}$ $=root(3)2^(3)*root(3)(2^4*x^2*y^7)$
$= \sqrt[3]{2^{3} \cdot 2^{4} \cdot x^{2} \cdot y^{7}}$ $=root(3)(2^3*2^4*x^2*y^7)$
$= \sqrt[3]{2^{7} x^{2} y^{7}}$ $=root(3)(2^7x^2y^7)$

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

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How do you simplify $2 \sqrt[3]{16 x^{2} y^{7}}$ $2\root[ 3] { 16x ^ { 2} y ^ { 7} }$ ?

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Explanation:

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How do you simplify $2 \sqrt[3]{16 x^{2} y^{7}}$ $2\root[ 3] { 16x ^ { 2} y ^ { 7} }$ ?