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Question-159170

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Question Number 159170 by mnjuly1970 last updated on 13/Nov/21
Commented by cortano last updated on 13/Nov/21
⇒(x)^(1/3) +((a−x))^(1/3) −((2a−x))^(1/3) = 0 ⇒x+a−x−2a+x = −3((x(a−x)(2a−x)))^(1/3) ⇒x−a = −3((x(a−x)(2a−x)))^(1/3) ⇒x^3 −3ax^2 +3a^2 x−a^3 =−27(ax−x^2 )(2a−x) ⇒x^3 −3ax^2 +3a^2 x−a^3 =−27(2a^2 x−3ax^2 +x^3 ) ⇒28x^3 −84ax^2 +57a^2 x−a^3 = 0 { (α),(β),(γ) :} ⇒α×β×γ = (a^3 /(28)) = (2/7) ⇒a^3 = 8 ; a=2
$$\Rightarrow\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow{x}+{a}−{x}−\mathrm{2}{a}+{x}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}−{a}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {x}−{a}^{\mathrm{3}} =−\mathrm{27}\left({ax}−{x}^{\mathrm{2}} \right)\left(\mathrm{2}{a}−{x}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {x}−{a}^{\mathrm{3}} =−\mathrm{27}\left(\mathrm{2}{a}^{\mathrm{2}} {x}−\mathrm{3}{ax}^{\mathrm{2}} +{x}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\mathrm{28}{x}^{\mathrm{3}} −\mathrm{84}{ax}^{\mathrm{2}} +\mathrm{57}{a}^{\mathrm{2}} {x}−{a}^{\mathrm{3}} \:=\:\mathrm{0}\:\begin{cases}{\alpha}\\{\beta}\\{\gamma}\end{cases} \\ $$$$\Rightarrow\alpha×\beta×\gamma\:=\:\frac{{a}^{\mathrm{3}} }{\mathrm{28}}\:=\:\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\Rightarrow{a}^{\mathrm{3}} \:=\:\mathrm{8}\:;\:{a}=\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Nov/21
(x)^(1/3) +((a−x))^(1/3) =((2a−x))^(1/3) ((x)^(1/3) +((a−x))^(1/3) )^3 =(((2a−x))^(1/3) )^3 x+(a−x)+3((x)^(1/3) ((a−x))^(1/3) )((x)^(1/3) +((a−x))^(1/3) ) =2a−x 3((x)^(1/3) ((a−x))^(1/3) )((2a−x))^(1/3) =a−x 27x(a−x)(2a−x)=(a−x)^3 (a−x)^3 −27x(a−x)(2a−x)=0 (a−x){(a−x)^2 −27x(2a−x)}=0 (a−x){28x^2 −56ax+a^2 }=0 x=a_(a=α (say)) ∣ 28x^2 −56ax+a^2 }=0_(βγ=a^2 /28) αβγ=a((a^2 /(28)))=(2/7)⇒a^3 =8⇒a=2
$$\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:=\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\: \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:\right)^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\:\right)^{\mathrm{3}} \\ $$$${x}+\left({a}−{x}\right)+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{x}}\:\sqrt[{\mathrm{3}}]{{a}−{x}}\:\right)\left(\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{a}−{x} \\ $$$$\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{x}}\:\sqrt[{\mathrm{3}}]{{a}−{x}}\:\right)\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\:\:={a}−{x} \\ $$$$\:\mathrm{27}{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)=\left({a}−{x}\right)^{\mathrm{3}} \\ $$$$\:\left({a}−{x}\right)^{\mathrm{3}} −\mathrm{27}{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)=\mathrm{0} \\ $$$$\left({a}−{x}\right)\left\{\left({a}−{x}\right)^{\mathrm{2}} −\mathrm{27}{x}\left(\mathrm{2}{a}−{x}\right)\right\}=\mathrm{0} \\ $$$$\left({a}−{x}\right)\left\{\mathrm{28}{x}^{\mathrm{2}} −\mathrm{56}{ax}+{a}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$$\left.\underset{{a}=\alpha\:\left({say}\right)} {{x}={a}}\mid\:\underset{\beta\gamma={a}^{\mathrm{2}} /\mathrm{28}} {\mathrm{28}{x}^{\mathrm{2}} −\mathrm{56}{ax}+{a}^{\mathrm{2}} \right\}=\mathrm{0}} \\ $$$$\alpha\beta\gamma={a}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{28}}\right)=\frac{\mathrm{2}}{\mathrm{7}}\Rightarrow{a}^{\mathrm{3}} =\mathrm{8}\Rightarrow{a}=\mathrm{2} \\ $$

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