Evaluate the cube root of 7 multiplied by the square root of 7 over the sixth root of 7 to the power of 5?

Solution:

Given the cube root of 7 multiplied by the square root of 7 over the sixth root of 7 to the power of 5

Let x = ((7)^1/3 × (7)^1/2) / (6√7)⁵

By using algebraic conditions

n√x = (x)^1/n

= (7)^1/3 × (7)^1/2 / (7)^5/6

Also from laws of exponents, we can write it as :

x^m × xⁿ = x^(m+n)

Therefore,

x = (7)^{1/3 + 1/2}/ (7)^5/6

⇒ 1/3 + 1/2 = 5/6

x = (7)^5/6 / (7)^5/6

By using quotient law of exponents, 

x^m/xⁿ = x^m-n

x = (7)^{5/6 - 5/6}

x= (7)⁰

But (a)⁰ = 1

x = 1

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Evaluate the cube root of 7 multiplied by the square root of 7 over the sixth root of 7 to the power of 5?

Summary:

The cube root of 7 multiplied by the square root of 7 over the sixth root of 7 to the power of 5 is 1.