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Mathematics

If x2+1x2x^2 + \dfrac{1}{x^2} = 7 and x≠ 0; find the value of :

7x3 + 8x - 7x3−8x\dfrac{7}{x^3} - \dfrac{8}{x}.

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Answer

By formula,

⇒(x−1x)2=x2+1x2−2⇒(x−1x)2=7−2⇒(x−1x)2=5⇒x−1x=±5.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm \sqrt{5}.

By formula,

⇒(x−1x)3=x3−1x3−3(x−1x)\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = x^3 - \dfrac{1}{x^3} - 3\Big(x - \dfrac{1}{x}\Big)

Substituting x−1x=5x - \dfrac{1}{x} = \sqrt{5} in above equation, we get :

⇒(5)3=x3−1x3−3×5⇒55=x3−1x3−35⇒x3−1x3=85.\Rightarrow (\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times \sqrt{5} \\[1em] \Rightarrow 5\sqrt{5} = x^3 - \dfrac{1}{x^3} - 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 8\sqrt{5}.

Substituting x−1x=−5x - \dfrac{1}{x} = -\sqrt{5} in above equation, we get :

⇒(−5)3=x3−1x3−3×−5⇒−55=x3−1x3+35⇒x3−1x3=−85.\Rightarrow (-\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times -\sqrt{5} \\[1em] \Rightarrow -5\sqrt{5} = x^3 - \dfrac{1}{x^3} + 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -8\sqrt{5}.

Solving given equation,

⇒7x3+8x−7x3−8x⇒7x3−7x3+8x−8x⇒7(x3−1x3)+8(x−1x)\Rightarrow 7x^3 + 8x - \dfrac{7}{x^3} - \dfrac{8}{x} \\[1em] \Rightarrow 7x^3 - \dfrac{7}{x^3} + 8x - \dfrac{8}{x} \\[1em] \Rightarrow 7\Big(x^3 - \dfrac{1}{x^3}\Big) + 8\Big(x - \dfrac{1}{x}\Big)

Substituting x3−1x3=85 and x−1x=5x^3 - \dfrac{1}{x^3} = 8\sqrt{5}\text{ and } x - \dfrac{1}{x} = \sqrt{5}, we get :

⇒7×85+8×5⇒565+85⇒645.\Rightarrow 7 \times 8\sqrt{5} + 8 \times \sqrt{5} \\[1em] \Rightarrow 56\sqrt{5} + 8\sqrt{5} \\[1em] \Rightarrow 64\sqrt{5}.

Substituting x3−1x3=−85 and x−1x=−5x^3 - \dfrac{1}{x^3} = -8\sqrt{5}\text{ and } x - \dfrac{1}{x} = -\sqrt{5}, we get :

⇒7×−85+8×−5⇒−565−85⇒−645.\Rightarrow 7 \times -8\sqrt{5} + 8 \times -\sqrt{5} \\[1em] \Rightarrow -56\sqrt{5} - 8\sqrt{5} \\[1em] \Rightarrow -64\sqrt{5}.

Hence, 7x3 + 8x - 7x3−8x=±645\dfrac{7}{x^3} - \dfrac{8}{x} = \pm 64\sqrt{5}.

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