If x = 15−x\dfrac{1}{5 - x}5−x1 and x ≠ 5, find : x3+1x3x^3 + \dfrac{1}{x^3}x3+x31.
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Given,
⇒x=15−x⇒x(5−x)=1⇒5x−x2=1⇒x2+1=5x⇒x2+1x=5xx⇒x+1x=5.\Rightarrow x = \dfrac{1}{5 - x} \\[1em] \Rightarrow x(5 - x) = 1 \\[1em] \Rightarrow 5x - x^2 = 1 \\[1em] \Rightarrow x^2 + 1 = 5x \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{5x}{x} \\[1em] \Rightarrow x + \dfrac{1}{x} = 5.⇒x=5−x1⇒x(5−x)=1⇒5x−x2=1⇒x2+1=5x⇒xx2+1=x5x⇒x+x1=5.
By formula,
⇒(x+1x)3=x3+1x3+3(x+1x)⇒53=x3+1x3+3×5⇒125=x3+1x3+15⇒x3+1x3=125−15=110.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 5^3 = x^3 + \dfrac{1}{x^3} + 3 \times 5 \\[1em] \Rightarrow 125 = x^3 + \dfrac{1}{x^3} + 15 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 125 - 15 = 110.⇒(x+x1)3=x3+x31+3(x+x1)⇒53=x3+x31+3×5⇒125=x3+x31+15⇒x3+x31=125−15=110.
Hence, x3+1x3=110.x^3 + \dfrac{1}{x^3} = 110.x3+x31=110.
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