Question

7\sqrt4{648x^{8}}

Explanation:

Step1: Factor 648 and simplify \(x^8\)

First, factor 648: \(648 = 81\times8 = 3^4\times8\). And for \(x^8\), using the property of exponents \((a^m)^n=a^{mn}\), we have \(x^8=(x^2)^4\). So the expression inside the fourth - root becomes \(648x^8 = 3^4\times8\times(x^2)^4\).
The original expression \(7\sqrt[4]{648x^8}\) can be rewritten as \(7\sqrt[4]{3^4\times8\times(x^2)^4}\).

Step2: Apply the fourth - root property \(\sqrt[4]{ab}=\sqrt[4]{a}\cdot\sqrt[4]{b}\)

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\) for even \(n\)), we get:
\(7\sqrt[4]{3^4\times8\times(x^2)^4}=7\times\sqrt[4]{3^4}\times\sqrt[4]{(x^2)^4}\times\sqrt[4]{8}\)
Since \(\sqrt[4]{3^4} = 3\) and \(\sqrt[4]{(x^2)^4}=x^2\) (for \(x\) in the domain where the radical is defined), the expression becomes \(7\times3\times x^2\times\sqrt[4]{8}\)

Step3: Simplify \(\sqrt[4]{8}\)

We know that \(8 = 2^3\), so \(\sqrt[4]{8}=\sqrt[4]{2^3}=2^{\frac{3}{4}}\). But we can also rewrite \(8\) as \(16\times\frac{1}{2}\), and \(\sqrt[4]{16\times\frac{1}{2}}=\sqrt[4]{16}\times\sqrt[4]{\frac{1}{2}} = 2\times2^{-\frac{1}{4}}=2^{\frac{3}{4}}\), but a better way is to factor 8 as \(2^3\) and then we can write the expression as:
\(7\times3\times x^2\times\sqrt[4]{8}=21x^2\sqrt[4]{8}\). However, we can also factor 648 as \(648 = 81\times8=3^4\times8\), and we can rewrite the original radical as:
\(7\sqrt[4]{648x^8}=7\sqrt[4]{3^4\times8\times x^8}=7\times3\times x^2\times\sqrt[4]{8}=21x^2\sqrt[4]{8}\). We can also simplify \(\sqrt[4]{8}\) as \(\sqrt[4]{2^3}=2^{\frac{3}{4}}\), but if we want to express it in terms of a perfect fourth - power factor, we note that \(8 = 16\times\frac{1}{2}\), so \(\sqrt[4]{8}=\sqrt[4]{16\times\frac{1}{2}} = 2\sqrt[4]{\frac{1}{2}}=2\times2^{-\frac{1}{4}}=2^{\frac{3}{4}}\). But another approach:
\(648=81\times8 = 3^4\times2^3\), so \(\sqrt[4]{648x^8}=\sqrt[4]{3^4\times2^3\times x^8}=3x^2\sqrt[4]{2^3}\)
Then the original expression \(7\sqrt[4]{648x^8}=7\times3x^2\sqrt[4]{8}=21x^2\sqrt[4]{8}\). If we want to rationalize or write it in a more simplified radical form, we can rewrite \(\sqrt[4]{8}\) as \(\sqrt[4]{16\times\frac{1}{2}} = 2\sqrt[4]{\frac{1}{2}}\), so \(21x^2\times2\sqrt[4]{\frac{1}{2}} = 42x^2\sqrt[4]{\frac{1}{2}}\), but the simplest form is \(21x^2\sqrt[4]{8}\) or we can factor 8 as \(2^3\) and write it as \(21x^2\times2^{\frac{3}{4}}\). However, maybe there was a typo and the number inside the radical is 64 instead of 648? If it was 64, then \(\sqrt[4]{64x^8}=\sqrt[4]{2^6x^8}=2^{\frac{6}{4}}x^2 = 2^{\frac{3}{2}}x^2=\sqrt{8}x^2\), but with 648, the above is the simplification.

Wait, let's re - do the factoring of 648 correctly. \(648\div2 = 324\), \(324\div2 = 162\), \(162\div2 = 81\), \(81 = 3^4\). So \(648=2^3\times3^4\). So \(\sqrt[4]{648x^8}=\sqrt[4]{2^3\times3^4\times x^8}=\sqrt[4]{3^4}\times\sqrt[4]{x^8}\times\sqrt[4]{2^3}=3\times x^2\times\sqrt[4]{8}\) (since \(\sqrt[4]{x^8}=x^{\frac{8}{4}} = x^2\) and \(\sqrt[4]{3^4}=3\)). Then multiplying by 7, we get \(7\times3\times x^2\times\sqrt[4]{8}=21x^2\sqrt[4]{8}\). We can also write \(\sqrt[4]{8}\) as \(2^{\frac{3}{4}}\) or as \(\sqrt[4]{8}\), and \(21x^2\sqrt[4]{8}\) can be written as \(21x^2\times2^{\frac{3}{4}}\) or we can rationalize the denominator of the radical part. But the main simplification steps are as above.

Answer:

\(21x^{2}\sqrt[4]{8}\) (or equivalent forms like \(21x^{2}\times2^{\frac{3}{4}}\))