If x−1x=5x - \dfrac{1}{x} = 5x−x1=5, find the value of (x3−1x3)\Big(x^3 - \dfrac{1}{x^3}\Big)(x3−x31).
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Given,
x−1x=5x - \dfrac{1}{x} = 5x−x1=5
Using identity,
⇒(x3−1x3)=(x−1x)3+3(x−1x)⇒(x3−1x3)=(5)3+3×5⇒(x3−1x3)=125+15⇒(x3−1x3)=140\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (5)^3 + 3 \times 5 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 125 + 15 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 140⇒(x3−x31)=(x−x1)3+3(x−x1)⇒(x3−x31)=(5)3+3×5⇒(x3−x31)=125+15⇒(x3−x31)=140
Hence, x3−1x3=140.x^3 - \dfrac{1}{x^3} = 140.x3−x31=140.
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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).
If x−2x=6x - \dfrac{2}{x} = 6x−x2=6, find the value of (x3−8x3)\Big(x^3 - \dfrac{8}{x^3}\Big)(x3−x38).
If x+1x=4x + \dfrac{1}{x} = 4x+x1=4, find the values of:
(i) (x3+1x3)\Big(x^3 + \dfrac{1}{x^3}\Big)(x3+x31)
(ii) (x−1x)\Big(x - \dfrac{1}{x}\Big)(x−x1)
(iii) (x3−1x3)\Big(x^3 - \dfrac{1}{x^3}\Big)(x3−x31)
