Answer:
[tex]\frac{\sqrt[3]{16y^4}}{x^2}[/tex]
Step-by-step explanation:
The options are missing; However, I'll simplify the given expression.
Given
[tex]\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }[/tex]
Required
Write Equivalent Expression
To solve this expression, we'll make use of laws of indices throughout.
From laws of indices [tex]\sqrt[n]{a} = a^{\frac{1}{n}}[/tex]
So,
[tex]\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }[/tex] gives
[tex]\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}[/tex]
Also from laws of indices
[tex](ab)^n = a^nb^n[/tex]
So, the above expression can be further simplified to
[tex]\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}[/tex]
Multiply the exponents gives
[tex]\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}[/tex]
Substitute [tex]2^5[/tex] for 32
[tex]\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}[/tex]
[tex]\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}[/tex]
From laws of indices
[tex]\frac{a^m}{a^n} = a^{m-n}[/tex]
This law can be applied to the expression above;
[tex]\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}[/tex] becomes
[tex]2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}[/tex]
Solve exponents
[tex]2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}[/tex]
[tex]2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}[/tex]
From laws of indices,
[tex]a^{-n} = \frac{1}{a^n}[/tex]; So,
[tex]2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}[/tex] gives
[tex]\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}[/tex]
The expression at the numerator can be combined to give
[tex]\frac{(2y)^{\frac{4}{3}}}{x^2}[/tex]
Lastly, From laws of indices,
[tex]a^{\frac{m}{n} = \sqrt[n]{a^m}[/tex]; So,
[tex]\frac{(2y)^{\frac{4}{3}}}{x^2}[/tex] becomes
[tex]\frac{\sqrt[3]{(2y)}^{4}}{x^2}[/tex]
[tex]\frac{\sqrt[3]{16y^4}}{x^2}[/tex]
Hence,
[tex]\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }[/tex] is equivalent to [tex]\frac{\sqrt[3]{16y^4}}{x^2}[/tex]
Answer is = [tex]\frac{2y}{x^{2}}\sqrt[3]{2y}[/tex]
We first take them all under the cube root sign, then simplify the inside.
We'll use the properties of exponents to simplify.
[tex]\frac{\sqrt[3]{32x^{3}y^{6}}}{\sqrt[3]{2x^{9}y^{2}}}\\=\sqrt[3]{\frac{32x^{3}y^{6}}{2x^{9}y^{2}}}\\=\sqrt[3]{\frac{16y^{\left(6-2\right)}}{x^{\left(9-3\right)}}}\\=\sqrt[3]{\frac{16y^{4}}{x^{6}}}\\=\sqrt[3]{\frac{2^{4}y^{4}}{x^{6}}}\\=\sqrt[3]{\frac{2^{3}y^{3}\cdot2y}{\left(x^{2}\right)^{3}}}\\=\frac{2y}{x^{2}}\sqrt[3]{2y}[/tex]
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