Thread: Cubed Root of the Square root of 27 = square root of 3? Why

1. Cubed Root of the Square root of 27 = square root of 3? Why

Can someone ask me how this problem is solved.

The cubed root of the square root of 27. Why does it equal the square root of 3?

2. Re: Cubed Root of the Square root of 27 = square root of 3? Why

Originally Posted by ttryon001
Can someone ask me how this problem is solved.

The cubed root of the square root of 27. Why does it equal the square root of 3?
First, $\sqrt[3]{\sqrt{27}}=(27)^{\frac{1}{6}}=$$(27^{\frac{1}{3}} )^{\frac{1}{2}}=3^{\frac{1}{2}}=\sqrt3$

3. Re: Cubed Root of the Square root of 27 = square root of 3? Why

$\sqrt[3]{\sqrt[2]{27}} =\sqrt[2]{\sqrt[3]{27}} = (27^{\frac{1}{3}} ) ^{ \frac{1}{2}} = 3^{\frac{1}{2}} = \sqrt[2]{2} = \sqrt2$

We are simply manipulating the expression. Since the properties of the exponents are commutative, we can take the cube root of 27 first which is 3 and then take the square root.

4. Re: Cubed Root of the Square root of 27 = square root of 3? Why

Originally Posted by ttryon001
Can someone ask me how this problem is solved.

The cubed root of the square root of 27. Why does it equal the square root of 3?
Here is another way

$\sqrt[3]{\sqrt{27}} = \sqrt[3]{\sqrt{9 * 3}} = \sqrt[3]{\sqrt{9} * \sqrt{3}} = \sqrt[3]{3 * \sqrt{3}} =$

$\sqrt[3]{\sqrt{3} * \sqrt{3} * \sqrt{3}} = \sqrt[3]{\left(\sqrt{3}\right)^3} = \sqrt{3}.$

which equation justifies why 27

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