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To simplify the given expression [tex]\(\frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}}\)[/tex], we will break it down step by step.
1. Simplify the Cube Roots Separately:
First, let's simplify the numerators and denominators inside the cube roots.
The numerator is [tex]\(\sqrt[3]{32 x^3 y^6}\)[/tex]:
[tex]\[ \sqrt[3]{32 x^3 y^6} = 32^{1/3} \cdot (x^3)^{1/3} \cdot (y^6)^{1/3} \][/tex]
Simplifying each term individually:
[tex]\[ 32^{1/3} = 2^{5/3} = 2 \cdot 2^{2/3} \][/tex]
[tex]\[ (x^3)^{1/3} = x \][/tex]
[tex]\[ (y^6)^{1/3} = y^2 \][/tex]
So,
[tex]\[ \sqrt[3]{32 x^3 y^6} = 2 \cdot x \cdot y^2 \cdot 2^{2/3} = 2 \cdot 2^{2/3} \cdot x \cdot y^2 \][/tex]
Next, let's simplify the denominator [tex]\(\sqrt[3]{2 x^9 y^2}\)[/tex]:
[tex]\[ \sqrt[3]{2 x^9 y^2} = (2)^{1/3} \cdot (x^9)^{1/3} \cdot (y^2)^{1/3} \][/tex]
Simplifying each term individually:
[tex]\[ 2^{1/3} \][/tex]
[tex]\[ (x^9)^{1/3} = x^3 \][/tex]
[tex]\[ (y^2)^{1/3} = y^{2/3} \][/tex]
So,
[tex]\[ \sqrt[3]{2 x^9 y^2} = 2^{1/3} \cdot x^3 \cdot y^{2/3} \][/tex]
2. Combine the Simplified Forms:
Now we put the simplified numerator and denominator together:
[tex]\[ \frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}} = \frac{2 \cdot 2^{2/3} \cdot x \cdot y^2}{2^{1/3} \cdot x^3 \cdot y^{2/3}} \][/tex]
3. Simplify the Fraction:
We can simplify this by canceling like terms and combining the powers of 2:
[tex]\[ \frac{2 \cdot 2^{2/3}}{2^{1/3}} = 2^{1 + 2/3 - 1/3} = 2^{4/3} \][/tex]
For the [tex]\(x\)[/tex] terms:
[tex]\[ \frac{x}{x^3} = x^{1-3} = x^{-2} \][/tex]
For the [tex]\(y\)[/tex] terms:
[tex]\[ \frac{y^2}{y^{2/3}} = y^{2 - 2/3} = y^{4/3} \][/tex]
Thus, the simplified form is:
[tex]\[ 2^{4/3} \cdot x^{-2} \cdot y^{4/3} \][/tex]
But [tex]\(2^{4/3} = (2^{1/3})^4\)[/tex], simplifying it separately:
[tex]\[ (2^{1/3})^4 = 2^{4/3} \][/tex]
Hence, the expression becomes:
[tex]\[ \frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}} = 2^{4/3} \cdot x^{-2} \cdot y^{4/3} \][/tex]
Therefore, the equivalent simplified form of [tex]\(\frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}}\)[/tex] is:
[tex]\[ 2^{4/3} \cdot \frac{y^{4/3}}{x^2} \][/tex]
1. Simplify the Cube Roots Separately:
First, let's simplify the numerators and denominators inside the cube roots.
The numerator is [tex]\(\sqrt[3]{32 x^3 y^6}\)[/tex]:
[tex]\[ \sqrt[3]{32 x^3 y^6} = 32^{1/3} \cdot (x^3)^{1/3} \cdot (y^6)^{1/3} \][/tex]
Simplifying each term individually:
[tex]\[ 32^{1/3} = 2^{5/3} = 2 \cdot 2^{2/3} \][/tex]
[tex]\[ (x^3)^{1/3} = x \][/tex]
[tex]\[ (y^6)^{1/3} = y^2 \][/tex]
So,
[tex]\[ \sqrt[3]{32 x^3 y^6} = 2 \cdot x \cdot y^2 \cdot 2^{2/3} = 2 \cdot 2^{2/3} \cdot x \cdot y^2 \][/tex]
Next, let's simplify the denominator [tex]\(\sqrt[3]{2 x^9 y^2}\)[/tex]:
[tex]\[ \sqrt[3]{2 x^9 y^2} = (2)^{1/3} \cdot (x^9)^{1/3} \cdot (y^2)^{1/3} \][/tex]
Simplifying each term individually:
[tex]\[ 2^{1/3} \][/tex]
[tex]\[ (x^9)^{1/3} = x^3 \][/tex]
[tex]\[ (y^2)^{1/3} = y^{2/3} \][/tex]
So,
[tex]\[ \sqrt[3]{2 x^9 y^2} = 2^{1/3} \cdot x^3 \cdot y^{2/3} \][/tex]
2. Combine the Simplified Forms:
Now we put the simplified numerator and denominator together:
[tex]\[ \frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}} = \frac{2 \cdot 2^{2/3} \cdot x \cdot y^2}{2^{1/3} \cdot x^3 \cdot y^{2/3}} \][/tex]
3. Simplify the Fraction:
We can simplify this by canceling like terms and combining the powers of 2:
[tex]\[ \frac{2 \cdot 2^{2/3}}{2^{1/3}} = 2^{1 + 2/3 - 1/3} = 2^{4/3} \][/tex]
For the [tex]\(x\)[/tex] terms:
[tex]\[ \frac{x}{x^3} = x^{1-3} = x^{-2} \][/tex]
For the [tex]\(y\)[/tex] terms:
[tex]\[ \frac{y^2}{y^{2/3}} = y^{2 - 2/3} = y^{4/3} \][/tex]
Thus, the simplified form is:
[tex]\[ 2^{4/3} \cdot x^{-2} \cdot y^{4/3} \][/tex]
But [tex]\(2^{4/3} = (2^{1/3})^4\)[/tex], simplifying it separately:
[tex]\[ (2^{1/3})^4 = 2^{4/3} \][/tex]
Hence, the expression becomes:
[tex]\[ \frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}} = 2^{4/3} \cdot x^{-2} \cdot y^{4/3} \][/tex]
Therefore, the equivalent simplified form of [tex]\(\frac{\sqrt[3]{32 x^3 y^6}}{\sqrt[3]{2 x^9 y^2}}\)[/tex] is:
[tex]\[ 2^{4/3} \cdot \frac{y^{4/3}}{x^2} \][/tex]
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