Question

rewrite the following expression in simplest radical form given $x > 0$.
\\(\sqrt3{216x^2} \cdot \sqrt4{256x}\\)

Explanation:

Step1: Simplify each radical separately

First, simplify \(\sqrt[3]{216x^{2}}\). We know that \(216 = 6^{3}\), so \(\sqrt[3]{216x^{2}}=\sqrt[3]{6^{3}x^{2}} = 6\sqrt[3]{x^{2}}\) (since \(\sqrt[3]{a^{3}}=a\) for any real number \(a\)).

Next, simplify \(\sqrt[4]{256x}\). We know that \(256 = 4^{4}\) (or \(256=2^{8}\), but \(4^{4}=256\) is more straightforward here), so \(\sqrt[4]{256x}=\sqrt[4]{4^{4}x}=4\sqrt[4]{x}\) (since \(\sqrt[4]{a^{4}} = a\) for \(a\geq0\), and here \(x>0\) so we can apply this).

Step2: Multiply the simplified radicals

Now, multiply \(6\sqrt[3]{x^{2}}\) and \(4\sqrt[4]{x}\). First, multiply the coefficients: \(6\times4 = 24\).

Then, we need to multiply the radicals \(\sqrt[3]{x^{2}}\) and \(\sqrt[4]{x}\). To multiply radicals with different indices, we can use the property \(a^{m}\cdot a^{n}=a^{m + n}\) by converting them to exponential form. Recall that \(\sqrt[n]{a}=a^{\frac{1}{n}}\), so \(\sqrt[3]{x^{2}}=x^{\frac{2}{3}}\) and \(\sqrt[4]{x}=x^{\frac{1}{4}}\).

Now, multiply \(x^{\frac{2}{3}}\cdot x^{\frac{1}{4}}\). Using the exponent rule \(a^{m}\cdot a^{n}=a^{m + n}\), we add the exponents: \(\frac{2}{3}+\frac{1}{4}=\frac{8 + 3}{12}=\frac{11}{12}\). So \(x^{\frac{2}{3}}\cdot x^{\frac{1}{4}}=x^{\frac{11}{12}}\).

But we want to write this back in radical form. Since \(x^{\frac{11}{12}}=\sqrt[12]{x^{11}}\) (because \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\)).

Wait, but maybe we made a mistake here. Let's check again. Wait, actually, when multiplying \(\sqrt[3]{x^{2}}\) and \(\sqrt[4]{x}\), another way is to find a common index. The least common multiple of 3 and 4 is 12. So:

\(\sqrt[3]{x^{2}}=x^{\frac{2}{3}}=x^{\frac{8}{12}}=\sqrt[12]{x^{8}}\)

\(\sqrt[4]{x}=x^{\frac{1}{4}}=x^{\frac{3}{12}}=\sqrt[12]{x^{3}}\)

Then, \(\sqrt[12]{x^{8}}\cdot\sqrt[12]{x^{3}}=\sqrt[12]{x^{8 + 3}}=\sqrt[12]{x^{11}}\)

So putting it all together, the product is \(24\sqrt[12]{x^{11}}\)? Wait, no, wait. Wait, actually, let's re - examine the initial simplification.

Wait, maybe we should simplify the radicals first by factoring the radicands into perfect powers for their respective indices.

Wait, let's start over.

First, \(\sqrt[3]{216x^{2}}\): \(216 = 6^{3}\), so \(\sqrt[3]{216x^{2}}=6\sqrt[3]{x^{2}}\) (correct).

\(\sqrt[4]{256x}\): \(256 = 2^{8}\), so \(\sqrt[4]{256x}=\sqrt[4]{2^{8}x}=2^{2}\sqrt[4]{x}=4\sqrt[4]{x}\) (since \(2^{8}=(2^{2})^{4}=4^{4}\), so \(\sqrt[4]{4^{4}x}=4\sqrt[4]{x}\), correct).

Now, multiplying \(6\sqrt[3]{x^{2}}\times4\sqrt[4]{x}=24\sqrt[3]{x^{2}}\sqrt[4]{x}\). Now, to combine these radicals, we use the formula \(\sqrt[n]{a}\cdot\sqrt[m]{b}=\sqrt[nm]{a^{m}b^{n}}\) (when \(n\) and \(m\) are positive integers).

So for \(\sqrt[3]{x^{2}}\) and \(\sqrt[4]{x}\), \(n = 3\), \(m = 4\), \(a=x^{2}\), \(b = x\). Then:

\(\sqrt[3]{x^{2}}\cdot\sqrt[4]{x}=\sqrt[12]{(x^{2})^{4}\cdot x^{3}}=\sqrt[12]{x^{8}\cdot x^{3}}=\sqrt[12]{x^{11}}\)

So the entire expression is \(24\sqrt[12]{x^{11}}\)? Wait, no, wait. Wait, let's check the exponents again.

Wait, \((x^{2})^{4}=x^{8}\) and \(x^{3}=x^{3}\), so \(x^{8}\cdot x^{3}=x^{11}\), so \(\sqrt[12]{x^{11}}\). So the product is \(24\sqrt[12]{x^{11}}\). But let's verify with another approach.

Alternatively, let's convert both radicals to exponent form first:

\(\sqrt[3]{216x^{2}}=(216x^{2})^{\frac{1}{3}}=(6^{3}x^{2})^{\frac{1}{3}}=6^{3\times\frac{1}{3}}x^{2\times\frac{1}{3}}=6x^{\frac{2}{3}}\)

\(\sqrt[4]{256x}=(256x)^{\frac{1}{4}}=(4^{4}x)^{\frac{1}{4}}=4^{4\times\frac{1}{4}}x^{\frac{1}{4}}=4x^{\frac{1}{4}}\)

Now, multiply them: \(6x^{\frac{2}{3}}\times4x^{\frac{1}{4}}=(6\times4)x^{\frac{2}{3}+\frac{1}{4}} = 24x^{\frac{8 + 3}{12}}=24x^{\frac{11}{12}}\)

And \(x^{\frac{11}{12}}=\sqrt[12]{x^{11}}\), so the expression in simplest radical form is \(24\sqrt[12]{x^{11}}\). Wait, but let's check if we can simplify the radicals further.

Wait, is there a mistake? Let's check the initial simplification of the radicals.

Wait, \(216 = 6^{3}\), so \(\sqrt[3]{216x^{2}} = 6\sqrt[3]{x^{2}}\) (correct). \(256=4^{4}\), so \(\sqrt[4]{256x}=4\sqrt[4]{x}\) (correct). Then multiplying the coefficients: \(6\times4 = 24\). Then for the variables: \(\sqrt[3]{x^{2}}\times\sqrt[4]{x}\). To combine these, we use the property of exponents for radicals: \(\sqrt[n]{a}\times\sqrt[m]{b}=\sqrt[nm]{a^{m}b^{n}}\). So \(n = 3\), \(m = 4\), \(a=x^{2}\), \(b=x\). Then \(a^{m}=(x^{2})^{4}=x^{8}\), \(b^{n}=x^{3}\), so \(a^{m}b^{n}=x^{8 + 3}=x^{11}\), and the index is \(nm = 12\), so \(\sqrt[12]{x^{11}}\). So the product is \(24\sqrt[12]{x^{11}}\).

Wait, but let's check with \(x = 1\) to verify.

If \(x = 1\), the original expression is \(\sqrt[3]{216\times1^{2}}\times\sqrt[4]{256\times1}=\sqrt[3]{216}\times\sqrt[4]{256}=6\times4 = 24\).

Now, the simplified expression when \(x = 1\) is \(24\sqrt[12]{1^{11}}=24\times1 = 24\), which matches. So that's correct.

Another check: let \(x = 16\). Original expression: \(\sqrt[3]{216\times16^{2}}\times\sqrt[4]{256\times16}\). \(16^{2}=256\), so \(\sqrt[3]{216\times256}=\sqrt[3]{55296}\). Let's calculate \(38^{3}=54872\), \(39^{3}=59319\), so that's not a perfect cube. Wait, no, wait, we made a mistake. Wait, no, when \(x = 16\), the original expression is \(\sqrt[3]{216\times16^{2}}\times\sqrt[4]{256\times16}=\sqrt[3]{216\times256}\times\sqrt[4]{4096}\). \(\sqrt[4]{4096}=8\) (since \(8^{4}=4096\)). \(\sqrt[3]{216\times256}=\sqrt[3]{55296}\). Let's factor 55296: \(55296\div6 = 9216\), \(9216 = 2^{10}\times3^{2}\)? Wait, no, \(216 = 6^{3}\), \(256 = 2^{8}\), so \(216\times256=6^{3}\times2^{8}\). Then \(\sqrt[3]{6^{3}\times2^{8}}=6\times2^{\frac{8}{3}}=6\times2^{2+\frac{2}{3}}=6\times4\times2^{\frac{2}{3}}=24\times\sqrt[3]{4}\). And the other part is \(\sqrt[4]{256\times16}=\sqrt[4]{2^{8}\times2^{4}}=\sqrt[4]{2^{12}}=2^{3}=8\). Wait, this is a contradiction. Wait, what's wrong here?

Ah! Wait a minute, I see the mistake. When we simplified \(\sqrt[4]{256x}\), if \(x = 16\), then \(\sqrt[4]{256\times16}=\sqrt[4]{4096}=\sqrt[4]{8^{4}} = 8\), which is correct. But when we simplified \(\sqrt[3]{216x^{2}}\) with \(x = 16\), \(\sqrt[3]{216\times16^{2}}=\sqrt[3]{216\times256}=\sqrt[3]{6^{3}\times2^{8}}=6\times2^{\frac{8}{3}}=6\times2^{2+\frac{2}{3}}=6\times4\times\sqrt[3]{4}=24\sqrt[3]{4}\). Then multiplying \(24\sqrt[3]{4}\times8 = 192\sqrt[3]{4}\). Now, let's check the simplified expression we had earlier: \(24\sqrt[12]{x^{11}}\) with \(x = 16\): \(24\sqrt[12]{16^{11}}\). \(16 = 2^{4}\), so \(16^{11}=2^{44}\), so \(\sqrt[12]{2^{44}}=2^{\frac{44}{12}}=2^{\frac{11}{3}}=2^{3+\frac{2}{3}}=8\sqrt[3]{4}\). Then \(24\times8\sqrt[3]{4}=192\sqrt[3]{4}\), which matches. And the original expression: \(\sqrt[3]{216\times16^{2}}\times\sqrt[4]{256\times16}=\sqrt[3]{55296}\times\sqrt[4]{4096}\). Let's calculate \(\sqrt[3]{55296}\): \(38^{3}=54872\), \(39^{3}=59319\), so \(\sqrt[3]{55296}\approx38.1\), and \(\sqrt[4]{4096}=8\), so \(38.1\times8\approx304.8\)? Wait, no, that's not right. Wait, I miscalculated \(216\times16^{2}\). \(16^{2}=256\), \(216\times256 = 216\times256\). Let's calculate \(200\times256=51200\), \(16\times256 = 4096\), so \(51200+4096 = 55296\). Now, \(\sqrt[3]{55296}\): let's factor 55296. Divide by 6: \(55296\div6 = 9216\). \(9216\div6 = 1536\). \(1536\div6 = 256\). \(256\div6\) is not an integer. Wait, 55296 = \(6^{3}\times2^{8}\) (since \(6^{3}=216\), \(2^{8}=256\), \(216\times256 = 55296\)). So \(\sqrt[3]{6^{3}\times2^{8}}=6\times2^{\frac{8}{3}}=6\times2^{2+\frac{2}{3}}=6\times4\times\sqrt[3]{4}=24\sqrt[3]{4}\approx24\times1.5874 = 38.0976\). And \(\sqrt[4]{256\times16}=\sqrt[4]{4096}=8\) (since \(8^{4}=4096\)). Then \(38.0976\times8\approx304.78\). Now, the simplified expression \(24\sqrt[12]{16^{11}}\): \(16^{11}=(2^{4})^{11}=2^{44}\), \(\sqrt[12]{2^{44}}=2^{\frac{44}{12}}=2^{\frac{11}{3}}=2^{3+\frac{2}{3}}=8\sqrt[3]{4}\approx8\times1.5874 = 12.699\). Then \(24\times12.699\approx304.78\), which matches. And the original calculation when I thought \(\sqrt[4]{256\times16}=8\) is correct, and \(\sqrt[3]{216\times16^{2}}\approx38.0976\), so their product is approximately \(304.78\), which matches the simplified expression. So my initial simplification was correct.

So going back, the steps are:

1. Simplify each radical:
- \(\sqrt[3]{216x^{2}}=6\sqrt[3]{x^{2}}\) (since \(216 = 6^{3}\))
- \(\sqrt[4]{256x}=4\sqrt[4]{x}\) (since \(256 = 4^{4}\))
2. Multiply the coefficients: \(6\times4 = 24\)
3. Multiply the variable - part radicals by converting to exponential form, adding exponents, and converting back to radical form:
- \(\sqrt[3]{x^{2}}=x^{\frac{2}{3}}\), \(\sqrt[4]{x}=x^{\frac{1}{4}}\)
- \(x^{\frac{2}{3}}\cdot x^{\frac{1}{4}}=x^{\frac{2}{3}+\frac{1}{4}}=x^{\frac{8 + 3}{12}}=x^{\frac{11}{12}}=\sqrt[12]{x^{11}}\)
4. Combine the coefficient and the radical: \(24\sqrt[12]{x^{11}}\)

Answer:

\(24\sqrt[12]{x^{11}}\)