# Radical expression by factoring out largest perfect n-th powerGiven: \sq

Radical expression by factoring out largest perfect n-th power
Given: $\sqrt[4]{64}$

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Calculation:
Step 1: Split into factors
$\sqrt[4]{64}=\sqrt[4]{2×2×2×2×2×2}$
$=\sqrt[4]{\left(2×2×2×2\right)×2×2}$
$=\sqrt[4]{\left(2{\right)}^{4}×2×2}$
$=\sqrt[4]{\left(2{\right)}^{4}×4}$
2 is the perfect 4th power factor
Step 2: Using product property $\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{b}$
$K\sqrt[4]{\left(2{\right)}^{4}×4}=\sqrt[4]{\left(2{\right)}^{4}}\cdot \sqrt[4]{4}$
Step 3: Using rule $\sqrt[a]{x}={x}^{\frac{1}{a}}$ to take the perfect 4th power factor out of the radical.
$\sqrt[4]{\left(2{\right)}^{4}}={2}^{\frac{4}{4}}={2}^{1}=2$
Step 4: $\sqrt[4]{\left(2{\right)}^{4}}\cdot \sqrt[4]{4}=2\sqrt[4]{4}$
Therefore, we have
$\sqrt[4]{64}=2\sqrt[4]{4}$
Conclusion: $\sqrt[4]{64}=2\sqrt[4]{4}$