To find the irrational cube root square root of 32 and rewrite it as the product of an integer and another irrational square root in simplest form, we will follow these steps:
-
Find the square root of 32: \[ \sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2} \]
-
Find the cube root of the result: Now we need to find the cube root of \(4\sqrt{2}\): \[ \sqrt[3]{4\sqrt{2}} = \sqrt[3]{4} \cdot \sqrt[3]{\sqrt{2}} = \sqrt[3]{4} \cdot \sqrt[3]{2^{1/2}} = \sqrt[3]{4} \cdot 2^{1/6} \]
To simplify \(\sqrt[3]{4}\): \[ \sqrt[3]{4} = \sqrt[3]{2^2} = 2^{2/3} \]
Therefore: \[ \sqrt[3]{4\sqrt{2}} = 2^{2/3} \cdot 2^{1/6} \]
To add the exponents: \[ 2^{2/3 + 1/6} = 2^{4/6 + 1/6} = 2^{5/6} \]
-
Final result: The expression \(2^{5/6}\) can be rewritten as: \[ 2^{5/6} = 2^{4/6} \cdot 2^{1/6} = 2^{2/3} \cdot 2^{1/6} \]
To express it in the form of an integer times an irrational square root: \[ 2^{2/3} = \sqrt[3]{2^4} = \sqrt[3]{16} \]
Thus: \[ \sqrt[3]{32} = 2^{5/6} = 2 \cdot 2^{1/6} = 2 \cdot \sqrt[3]{2} \]
Finally, the irrational cube root square root of 32 can be expressed as: \[ \sqrt[3]{\sqrt{32}} = 2 \cdot 2^{1/6} = 2\sqrt[3]{2} \]
So, in simplest form, we can express the irrational cube root square root of 32 as: \[ \sqrt[3]{\sqrt{32}} = 2 \cdot \sqrt[3]{2} \]