9 divided by radical 27a to the fifth power
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9/[(√27)^5(√a)^5]
= 3^2/[3^(15/2)(a^2√a)]
= 3^(-11/2)/(a^2√a)
or
= 1/[(3^5)(3^1/2)(a^2√a)]
= 1/[(243√3)(a^2√a)]
messy looking thing!
actually looks better in the original form of 9/(√27a)^5
To solve the expression (9 / √(27a)^5), let's break it down step by step.
First, simplify the expression inside the square root. We can simplify the radical by finding the perfect square factor of 27. In this case, 9 is a perfect square that evenly divides into 27.
√27 = √(9 * 3) = √9 * √3 = 3√3
Now our expression becomes:
(9 / 3√3)^5
Next, simplify the fraction by dividing both the numerator and denominator by 3.
(9 / 3√3) = 3 / √3
Now our expression becomes:
(3 / √3)^5
To simplify further, we'll rationalize the denominator. Multiply both the numerator and denominator by the conjugate of the denominator, which is √3.
(3 / √3)^5 * (√3 / √3) = 3√3 / 3
Simplify the fraction:
3√3 / 3 = √3
So, the simplified expression is √3.