Question

the radical expression below has a radical in the denominator. fill in the missing radicand to rationalize the denominator.
\\(\frac{1}{\sqrt4{216b}} = \frac{1}{\sqrt4{216b}} \cdot \frac{\sqrt4{?}}{\sqrt4{?}}\\)

Explanation:

Step1: Recall the rule for rationalizing a fourth root denominator.

To rationalize the denominator of a fourth root, we need to make the radicand a perfect fourth power. The denominator is \(\sqrt[4]{216b}\), and \(216 = 6^3\), so the radicand is \(6^3b\). We need to multiply by \(\sqrt[4]{6b^3}\) (wait, no, let's do it properly). The general rule is that for \(\sqrt[n]{x}\), we multiply numerator and denominator by \(\sqrt[n]{x^{n - 1}}\) to make the radicand \(x^n\). Here, \(n = 4\), and the radicand in the denominator is \(216b=6^3b\). So we need to find what to multiply so that the radicand becomes a perfect fourth power. Let's factor \(216b\): \(216 = 6^3=2^3\times3^3\), so \(216b = 2^3\times3^3\times b\). To make it a perfect fourth power, we need each exponent to be a multiple of 4. So for the exponents of 2, 3, and \(b\): we have 3, so we need \(4 - 3 = 1\) more of each? Wait, no. Wait, the radicand is \(216b = 6^3b\). Let's let the missing radicand be \(x\), so that \(216b\times x\) is a perfect fourth power. Let \(216b = 6^3b\), so we need \(6^3b\times x = 6^4b^4\)? No, wait, let's take the prime factors. \(216 = 2^3\times3^3\), so \(216b = 2^3\times3^3\times b^1\). To make each exponent a multiple of 4, we need \(2^{4 - 3}=2^1\), \(3^{4 - 3}=3^1\), and \(b^{4 - 1}=b^3\)? Wait, no, maybe I messed up. Wait, the index is 4, so we need the radicand in the denominator after multiplication to be a perfect fourth power. So the original denominator radicand is \(216b\), so we multiply numerator and denominator by \(\sqrt[4]{216b^{3}}\)? Wait, no, let's do it step by step. Let's denote the denominator as \(\sqrt[4]{216b}\). To rationalize, we multiply numerator and denominator by \(\sqrt[4]{(216b)^{3}}\)? No, wait, no. Wait, the formula for rationalizing a single radical in the denominator with index \(n\) is to multiply numerator and denominator by \(\sqrt[n]{x^{n - 1}}\) where \(x\) is the radicand. So here, \(n = 4\), \(x = 216b\), so we multiply by \(\sqrt[4]{(216b)^{3}}\)? Wait, no, that would make the radicand \((216b)^4\), which is a perfect fourth power, but that's too much. Wait, no, actually, the radicand in the denominator is \(216b = 6^3b\). So we need to find the smallest \(x\) such that \(6^3b\times x\) is a perfect fourth power. Let's let \(x = 6b^3\), then \(6^3b\times6b^3 = 6^{4}b^{4}=(6b)^4\), which is a perfect fourth power. Let's check: \(6^3\times6 = 6^{4}\), \(b\times b^3 = b^4\), so \(6^3b\times6b^3 = 6^4b^4=(6b)^4\), which is a perfect fourth power. So the missing radicand is \(6b^3\)? Wait, wait, let's check the original problem. The denominator is \(\sqrt[4]{216b}\), so when we multiply numerator and denominator by \(\sqrt[4]{6b^3}\), then the denominator becomes \(\sqrt[4]{216b\times6b^3}=\sqrt[4]{1296b^4}=\sqrt[4]{(6b)^4}=6b\), which is rational. Wait, \(216\times6 = 1296\), and \(b\times b^3 = b^4\), so \(216b\times6b^3 = 1296b^4=(6b)^4\), which is a perfect fourth power. So the missing radicand is \(6b^3\)? Wait, but let's see the problem: the numerator is 1, and we have \(\frac{1}{\sqrt[4]{216b}}=\frac{1}{\sqrt[4]{216b}}\cdot\frac{\sqrt[4]{?}}{\sqrt[4]{?}}\). So we need to find? such that \(216b\times?\) is a perfect fourth power. Let's factor \(216b\): \(216 = 6^3 = 2^3\times3^3\), so \(216b = 2^3\times3^3\times b\). To make it a perfect fourth power, we need to multiply by \(2^1\times3^1\times b^3\) (since 3 + 1 = 4 for exponents of 2 and 3, and 1 + 3 = 4 for exponent of b). So \(2\times3\times b^3 = 6b^3\). So the missing radicand is \(6b^3\). Wait, but let's check: \(216b\times6b^3 = 1296b^4 = (6b)^4\), which is a perfect fourth power. So yes, the missing radicand is \(6b^3\). Wait, but maybe I made a mistake. Let's re-express:

The denominator is \(\sqrt[4]{216b}\). To rationalize, we multiply numerator and denominator by \(\sqrt[4]{(216b)^{3}}\)? No, that would be \(\sqrt[4]{216^3b^3}\), but that's not necessary. Wait, the key is that for a fourth root, we need the radicand to be a perfect fourth power. So if the radicand is \(216b = 6^3b\), then we need to multiply by \(6b^3\) so that \(6^3b\times6b^3 = 6^4b^4 = (6b)^4\), which is a perfect fourth power. So the missing radicand is \(6b^3\).

Step2: Verify the multiplication.

Multiply the denominator's radicand: \(216b\times6b^3 = 1296b^4\). The fourth root of \(1296b^4\) is \(\sqrt[4]{(6b)^4}=6b\), which is rational. So the missing radicand is \(6b^3\).

Answer:

\(6b^3\)