Question

question
simplify \\(\sqrt4{1296x^8}\\) completely given \\(x > 0\\).

answer attempt 1 out of 5

Explanation:

Step1: Simplify the constant term

We know that \(1296 = 6^4\) because \(6\times6\times6\times6 = 1296\). So we can rewrite the fourth - root of the constant part as \(\sqrt[4]{6^4}\). By the property of \(n\) - th roots, \(\sqrt[n]{a^n}=a\) when \(n\) is even and \(a\geq0\). Here \(n = 4\) and \(6\gt0\), so \(\sqrt[4]{6^4}=6\).

Step2: Simplify the variable term

For the variable part \(x^8\), we can rewrite it as \((x^{2})^4\) (using the power - of - a - power rule \((a^m)^n=a^{mn}\), here \(m = 2\) and \(n = 4\), so \((x^{2})^4=x^{2\times4}=x^8\)). Then, by the property of \(n\) - th roots \(\sqrt[n]{a^n}=a\) (since \(x\gt0\), \(x^{2}\gt0\) and \(n = 4\) is even), we have \(\sqrt[4]{(x^{2})^4}=x^{2}\).

Step3: Combine the results

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}\) (where \(a = 6^4\) and \(b=(x^{2})^4\) and \(n = 4\)), we get \(\sqrt[4]{1296x^{8}}=\sqrt[4]{6^4\times(x^{2})^4}=\sqrt[4]{6^4}\times\sqrt[4]{(x^{2})^4}\). Substituting the values we found in Step 1 and Step 2, we have \(6\times x^{2}=6x^{2}\).

Answer:

\(6x^{2}\)