Adding and subtracting radicals $$\displaystyle{2}{x}\sqrt{{{3}{x}^{{2}}{y}}}\cdot{3}\sqrt{{{15}{x}{y}^{{3}}}}$$

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Leonard Stokes
Given
$$\displaystyle{2}{x}\sqrt{{{3}{x}^{{2}}{y}}}\cdot{3}\sqrt{{{15}{x}{y}^{{3}}}}$$
We have to solve for the simple form
$$\displaystyle{\left({2}{x}\right)}{\left({3}\right)}\cdot\sqrt{{{3}{x}^{{2}}{y}}}\cdot\sqrt{{{15}{x}{y}^{{3}}}}$$
$$\displaystyle={6}{x}\cdot\sqrt{{{\left({3}{x}^{{2}}{y}\right)}\cdot{\left({15}{x}{y}^{{3}}\right)}}}$$
$$\displaystyle={6}{x}\cdot\sqrt{{{\left({3}\right)}{\left({15}\right)}{\left({x}^{{2}}\right)}{\left({x}\right)}{\left({y}\right)}{\left({y}^{{3}}\right)}}}$$
$$\displaystyle={6}{x}\cdot\sqrt{{{45}\cdot{x}^{{{2}+{1}}}\cdot{y}^{{{1}+{3}}}}}$$
$$\displaystyle={6}{x}\cdot\sqrt{{{45}\cdot{x}^{{3}}\cdot{y}^{{4}}}}$$
$$\displaystyle={6}{x}\sqrt{{{\left({9}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}$$
$$\displaystyle={6}{x}\sqrt{{{\left({9}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}$$
$$\displaystyle={6}{x}\sqrt{{{\left({3}^{{2}}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}$$
$$\displaystyle={6}{x}\sqrt{{{\left({3}{y}^{{2}}\right)}^{{2}}\cdot{5}{x}^{{3}}}}$$
$$\displaystyle={6}{x}\cdot{\left({3}{y}^{{2}}\right)}\cdot\sqrt{{{5}{x}^{{3}}}}$$
$$\displaystyle={18}{x}{y}^{{2}}\cdot\sqrt{{{5}{x}^{{3}}}}$$