Algebra

5th_root(64x^10y^5)

Step-by-step solution with explanation

Final Answer

Step-by-step solution

Rewrite the expression clearly

We are finding the 5th root of the entire expression inside. This means we want something raised to the 5th power that equals the expression inside.

Break 64 into a perfect fifth power

64 is not a perfect 5th power, but we can split it into a part that is (2^5 = 32) and a leftover (2). So 64 = 32 · 2.

Rewrite all parts with 5th powers

We split 64 into 2^5 · 2, and keep x^10 and y^5 as they are. Now it's easy to see which parts come out of the radical.

Apply the 5th root to each factor

The 5th root of a product equals the product of the 5th roots. We separate each factor so we can simplify them one at a time.

Simplify each 5th root

, (since 10÷5=2), and . The stays inside because 2 is not a perfect 5th power.

Write the final simplified form

Multiply all the parts that came out of the radical together: 2, x², and y. The leftover ⁵√2 stays under the radical sign.

Understanding this problem

Learning Insight

The 5th root undoes raising to the 5th power. When you break each factor into groups of 5 exponents, those groups come out cleanly. Any leftover exponents (less than 5) must stay inside the radical. This is the same logic used for square roots, just with groups of 5 instead of 2.

Quick Tip

Divide each exponent by 5: the quotient is the outside power, and the remainder stays inside. For x^10: 10÷5 = 2 remainder 0, so x² comes out clean. For 2^6: 6÷5 = 1 remainder 1, so 2 comes out and 2^1 stays inside.

Common Mistake

Students often treat 64 as a perfect 5th power and write just 2 outside without keeping the leftover ⁵√2 inside. Remember: 2^5 = 32, not 64, so there IS a leftover factor of 2 that must stay under the radical.