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Question Number 64744 by mathmax by abdo last updated on 21/Jul/19

solve^3 (√(1+x))+^3 (√(1−x))=2

$${solve}\:^{\mathrm{3}} \sqrt{\mathrm{1}+{x}}+^{\mathrm{3}} \sqrt{\mathrm{1}−{x}}=\mathrm{2} \\ $$

Answered by MJS last updated on 21/Jul/19

a^(1/3) +b^(1/3) =c a+3a^(2/3) b^(1/3) +3a^(1/3) b^(2/3) +b=c^3 a+3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )+b=c^3 a+3a^(1/3) b^(1/3) c+b=c^3 3a^(1/3) b^(1/3) c=c^3 −a−b 27abc^3 =(c^3 −a−b)^3 216(1+x)(1−x)=(8−(1+x)−(1−x))^3 216(1−x^2 )=216 1−x^2 =1 x=0

$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c} \\ $$$${a}+\mathrm{3}{a}^{\frac{\mathrm{2}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}={c}^{\mathrm{3}} \\ $$$${a}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+{b}={c}^{\mathrm{3}} \\ $$$${a}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} {c}+{b}={c}^{\mathrm{3}} \\ $$$$\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} {c}={c}^{\mathrm{3}} −{a}−{b} \\ $$$$\mathrm{27}{abc}^{\mathrm{3}} =\left({c}^{\mathrm{3}} −{a}−{b}\right)^{\mathrm{3}} \\ $$$$\mathrm{216}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)=\left(\mathrm{8}−\left(\mathrm{1}+{x}\right)−\left(\mathrm{1}−{x}\right)\right)^{\mathrm{3}} \\ $$$$\mathrm{216}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{216} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}=\mathrm{0} \\ $$

Answered by behi83417@gmail.com last updated on 21/Jul/19

1+x=t^3 ,1−x=s^3 ⇒ { ((s+t=2)),((t^3 +s^3 =2)) :} ⇒(s+t)[(s+t)^2 −3st]=2 2(4−3st)=2⇒4−3st=1⇒st=1 z^2 −2z+1=0⇒z=1⇒ { ((1+x=1)),((1−x=1)) :}⇒x=0 .■

$$\mathrm{1}+\mathrm{x}=\mathrm{t}^{\mathrm{3}} ,\mathrm{1}−\mathrm{x}=\mathrm{s}^{\mathrm{3}} \Rightarrow\begin{cases}{\mathrm{s}+\mathrm{t}=\mathrm{2}}\\{\mathrm{t}^{\mathrm{3}} +\mathrm{s}^{\mathrm{3}} =\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{s}+\mathrm{t}\right)\left[\left(\mathrm{s}+\mathrm{t}\right)^{\mathrm{2}} −\mathrm{3st}\right]=\mathrm{2} \\ $$$$\mathrm{2}\left(\mathrm{4}−\mathrm{3st}\right)=\mathrm{2}\Rightarrow\mathrm{4}−\mathrm{3st}=\mathrm{1}\Rightarrow\mathrm{st}=\mathrm{1} \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{z}=\mathrm{1}\Rightarrow\begin{cases}{\mathrm{1}+\mathrm{x}=\mathrm{1}}\\{\mathrm{1}−\mathrm{x}=\mathrm{1}}\end{cases}\Rightarrow\mathrm{x}=\mathrm{0}\:.\blacksquare \\ $$

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