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If-x-x-1-1-find-x-1-3-1-x-1-3-




Question Number 89938 by jagoll last updated on 20/Apr/20
If x(x+1) = 1   find (x+1)^3 −(1/((x+1)^3 ))
$$\mathrm{If}\:\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{1}\: \\ $$$$\mathrm{find}\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by john santu last updated on 20/Apr/20
let x+1= t ⇒x = t−1  ⇒(t−1)t = 1 , t−1 = (1/t)  t−(1/t) = 1 ⇒(t−(1/t))^3 = 1   t^3 −3t+(3/t)−(1/t^3 ) = 1   t^3 −(1/t^3 ) −3(t−(1/t)) = 1  t^3 −(1/t^3 ) = 4 ⇒ (x+1)^3 −(1/((x+1)^3 )) = 4
$${let}\:{x}+\mathrm{1}=\:{t}\:\Rightarrow{x}\:=\:{t}−\mathrm{1} \\ $$$$\Rightarrow\left({t}−\mathrm{1}\right){t}\:=\:\mathrm{1}\:,\:{t}−\mathrm{1}\:=\:\frac{\mathrm{1}}{{t}} \\ $$$${t}−\frac{\mathrm{1}}{{t}}\:=\:\mathrm{1}\:\Rightarrow\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} =\:\mathrm{1}\: \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}+\frac{\mathrm{3}}{{t}}−\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:=\:\mathrm{1}\: \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:−\mathrm{3}\left({t}−\frac{\mathrm{1}}{{t}}\right)\:=\:\mathrm{1} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:=\:\mathrm{4}\:\Rightarrow\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\mathrm{4}\: \\ $$

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