If x+1x=3x + \dfrac{1}{x} = 3x+x1=3, find the value of (x3+1x3)\Big(x^3 + \dfrac{1}{x^3}\Big)(x3+x31).
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Given,
x+1x=3x + \dfrac{1}{x} = 3x+x1=3
Using identity,
⇒(x+1x)3=x3+1x3+3(x+1x)⇒33=x3+1x3+3×3⇒27=x3+1x3+9⇒x3+1x3=27−9=18.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 3^3 = x^3 + \dfrac{1}{x^3} + 3 \times 3 \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{x^3} + 9 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 27 - 9 = 18.⇒(x+x1)3=x3+x31+3(x+x1)⇒33=x3+x31+3×3⇒27=x3+x31+9⇒x3+x31=27−9=18.
Hence, x3+1x3=18.x^3 + \dfrac{1}{x^3} = 18.x3+x31=18.
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