Question

assuming ( x ) and ( y ) are both positive, write the following expression in simplest radical form. ( 3xsqrt{72x^{6}y^{3}} )

Explanation:

Step1: Factor the radicand

Factor \(72x^6y^3\) into perfect squares and remaining factors: \(72 = 36\times2\), \(x^6=(x^3)^2\), \(y^3 = y^2\times y\). So \(72x^6y^3=36\times2\times(x^3)^2\times y^2\times y\).

Step2: Simplify the square root

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a,b\geq0\)) and \(\sqrt{a^2}=a\) (\(a\geq0\)):
\(\sqrt{72x^6y^3}=\sqrt{36\times2\times(x^3)^2\times y^2\times y}=\sqrt{36}\cdot\sqrt{(x^3)^2}\cdot\sqrt{y^2}\cdot\sqrt{2y}=6x^3y\sqrt{2y}\)

Step3: Multiply by the coefficient outside

Multiply \(3x\) with \(6x^3y\sqrt{2y}\): \(3x\times6x^3y\sqrt{2y}=18x^{1 + 3}y\sqrt{2y}=18x^4y\sqrt{2y}\)

Answer:

\(18x^4y\sqrt{2y}\)