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Question Number 172513 by Mikenice last updated on 28/Jun/22

$$simplify$$$$\sqrt{a^{2}b+b^3+2{ab}^2}-\sqrt{a^{2}b+4b^3+4{ab}^3}$$

Commented by Rasheed.Sindhi last updated on 28/Jun/22

$$\mathcal{D}oyoumean:$$$$\sqrt{a^{2}b+b^3+2{ab}^2}-\sqrt{a^{2}b+4b^3+4{ab}^2}$$

Commented by Mikenice last updated on 28/Jun/22

$$yes$$

Answered by Rasheed.Sindhi last updated on 28/Jun/22

$$\sqrt{a^{2}b+b^3+2{ab}^2}-\sqrt{a^{2}b+4b^3+4{ab}^2}$$$$=\sqrt{b(a^2+b^2+2ab)}-\sqrt{b(a^2+4b^2+4ab)}$$$$=\sqrt{b{(a+b)}^2}-\sqrt{b{(a+2b)}^2}$$$$=\sqrt{b}((a+b)-(a+2b))$$$$=\sqrt{b}(a+b-a-2b))$$$$=-b\sqrt{b}$$

Answered by MJS_new last updated on 28/Jun/22

$$\sqrt{a^{2}b+b^3+2{ab}^2}=\mid a+b\mid \sqrt{b}$$$$\sqrt{a^{2}b+4b^3+4{ab}^2}=\mid a+2b\mid \sqrt{b}$$$$\mathrm{different}\mathrm{results}\mathrm{depending}\mathrm{on}a+b\mathrm{and}$$$$a+2b>=<0$$

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