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Question-145238

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Question Number 145238 by liberty last updated on 03/Jul/21
Answered by gsk2684 last updated on 03/Jul/21
x=1 or (1/3)
$$\mathrm{x}=\mathrm{1}\:\mathrm{or}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 03/Jul/21
(((x+1)^3 −(x−1)^3 )/((x+1)^2 −(x−1)^2 ))=2 (((x+1)/(2−x)))^(1/3) +(((2−x)/(x+1)))^(1/3) =? ((a^3 −b^3 )/(a^2 −b^2 ))=(((a−b)(a^2 +ab+b^2 ))/((a−b)(a+b))) =((a^2 +ab+b^2 )/(a+b)) (((x+1)^2 +(x+1)(x−1)+(x−1)^2 )/((x+1)+(x−1)))=2 x^2 +2x+1+x^2 −1+x^2 −2x+1=4x 3x^2 −4x+1=0 (x−1)(3x−1)=0 x=1 ∨ x=1/3 (a) x=1 (((x+1)/(2−x)))^(1/3) +(((2−x)/(x+1)))^(1/3) =(((1+1)/(2−1)))^(1/3) +(((2−1)/(1+1)))^(1/3) =(2)^(1/3) +(1/( (2)^(1/3) ))=2^(1/3) +2^(−1/3) (b)x=1/3 (((x+1)/(2−x)))^(1/3) +(((2−x)/(x+1)))^(1/3) =(((1/3+1)/(2−1/3)))^(1/3) +(((2−1/3)/(1/3+1)))^(1/3) (((4/3)/(5/3)))^(1/3) +(((5/3)/(4/3)))^(1/3) =((4/5))^(1/3) +((5/4))^(1/3)
$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{\mathrm{2}−{x}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}−{x}}{{x}+\mathrm{1}}}=? \\ $$$$\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)}{\left({a}−{b}\right)\left({a}+{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{{a}+{b}} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)+\left({x}−\mathrm{1}\right)}=\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\cancel{\mathrm{2}{x}}+\cancel{\mathrm{1}}+{x}^{\mathrm{2}} −\cancel{\mathrm{1}}+{x}^{\mathrm{2}} −\cancel{\mathrm{2}{x}}+\mathrm{1}=\mathrm{4}{x} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{1}/\mathrm{3} \\ $$$$\:\:\left(\boldsymbol{{a}}\right)\:{x}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{\mathrm{2}−{x}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}−{x}}{{x}+\mathrm{1}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}−\mathrm{1}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}−\mathrm{1}}{\mathrm{1}+\mathrm{1}}} \\ $$$$=\sqrt[{\mathrm{3}}]{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}=\mathrm{2}^{\mathrm{1}/\mathrm{3}} +\mathrm{2}^{−\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\left(\boldsymbol{{b}}\right){x}=\mathrm{1}/\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{\mathrm{2}−{x}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}−{x}}{{x}+\mathrm{1}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}/\mathrm{3}+\mathrm{1}}{\mathrm{2}−\mathrm{1}/\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}−\mathrm{1}/\mathrm{3}}{\mathrm{1}/\mathrm{3}+\mathrm{1}}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}/\mathrm{3}}{\mathrm{5}/\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}/\mathrm{3}}{\mathrm{4}/\mathrm{3}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$

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