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Question Number 145238 by liberty last updated on 03/Jul/21

Answered by gsk2684 last updated on 03/Jul/21

$$\mathrm{x}=1\mathrm{or}\frac{1}{3}$$

Answered by Rasheed.Sindhi last updated on 03/Jul/21

$$\frac{{(x+1)}^3-{(x-1)}^3}{{(x+1)}^2-{(x-1)}^2}=2$$$$\sqrt[3]{\frac{x+1}{2-x}}+\sqrt[3]{\frac{2-x}{x+1}}=?$$$$\frac{a^3-b^3}{a^2-b^2}=\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$$$$=\frac{a^2+ab+b^2}{a+b}$$$$\frac{{(x+1)}^2+(x+1)(x-1)+{(x-1)}^2}{(x+1)+(x-1)}=2$$$$x^2+\cancel{2x}+\cancel{1}+x^2-\cancel{1}+x^2-\cancel{2x}+1=4x$$$$3x^2-4x+1=0$$$$(x-1)(3x-1)=0$$$$x=1\vee x=1/3$$$$\left(\mathit{a}\right)x=1$$$$\sqrt[3]{\frac{x+1}{2-x}}+\sqrt[3]{\frac{2-x}{x+1}}=\sqrt[3]{\frac{1+1}{2-1}}+\sqrt[3]{\frac{2-1}{1+1}}$$$$=\sqrt[3]{2}+\frac{1}{\sqrt[3]{2}}=2^{1/3}+2^{-1/3}$$$$\left(\mathit{b}\right)x=1/3$$$$\sqrt[3]{\frac{x+1}{2-x}}+\sqrt[3]{\frac{2-x}{x+1}}=\sqrt[3]{\frac{1/3+1}{2-1/3}}+\sqrt[3]{\frac{2-1/3}{1/3+1}}$$$$\sqrt[3]{\frac{4/3}{5/3}}+\sqrt[3]{\frac{5/3}{4/3}}=\sqrt[3]{\frac{4}{5}}+\sqrt[3]{\frac{5}{4}}$$

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