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Question Number 44231 by Mawitea Cck last updated on 24/Sep/18

$$\mathrm{The}\mathrm{square}\mathrm{root}\mathrm{of}{x}^{m^2-n^2}\cdot x^{n^2+2mn}\cdot x^{n^2}\mathrm{is}$$

Answered by Joel578 last updated on 24/Sep/18

$$\sqrt{x^{(m^2-n^2)+(n^2+2mn)+n^2}}$$$$=\sqrt{x^{m^2+2mn+n^2}}$$$$=\sqrt{x^{{(m+n)}^2}}$$$$={\left(x^{{(m+n)}^2}\right)}^{\frac{1}{2}}$$$$=x^{m+n}$$

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