Question

simplify the expression.
\sqrt{54b^8}

Explanation:

Step1: Factor the radicand

Factor \(54b^8\) into perfect - square factors and non - perfect - square factors. We know that \(54 = 9\times6\) and \(b^8=(b^4)^2\). So, \(\sqrt{54b^8}=\sqrt{9\times6\times(b^4)^2}\).

Step2: Use the property of square roots \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\))

We can split the square root of the product into the product of square roots: \(\sqrt{9\times6\times(b^4)^2}=\sqrt{9}\times\sqrt{(b^4)^2}\times\sqrt{6}\).

Step3: Simplify the square roots of perfect squares

We know that \(\sqrt{9} = 3\) and \(\sqrt{(b^4)^2}=b^4\) (since \(b^8=(b^4)^2\) and for any real number \(x\), \(\sqrt{x^2}=\vert x\vert\), and if we assume \(b\) is a real number and in the context of simplifying algebraic expressions, we can write \(\sqrt{(b^4)^2}=b^4\) as \(b^4\) is non - negative for real \(b\) when considering the domain of the original expression (the expression under the square root must be non - negative, and \(b^8\) is always non - negative, and \(54\) is positive, so \(b\) can be any real number, and \(b^4\) is non - negative)). So, \(\sqrt{9}\times\sqrt{(b^4)^2}\times\sqrt{6}=3\times b^4\times\sqrt{6}\).

Step4: Combine the terms

Combining the terms, we get \(3b^4\sqrt{6}\).

Answer:

\(3b^4\sqrt{6}\)