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Mathematics

If x+1x=4x + \dfrac{1}{x} = 4, find the values of:

(i) (x3+1x3)\Big(x^3 + \dfrac{1}{x^3}\Big)

(ii) (x−1x)\Big(x - \dfrac{1}{x}\Big)

(iii) (x3−1x3)\Big(x^3 - \dfrac{1}{x^3}\Big)

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Answer

(i) Given,

x+1x=4x + \dfrac{1}{x} = 4

Using identity,

⇒(x+1x)3=x3+1x3+3(x+1x)⇒43=x3+1x3+3×4⇒64=x3+1x3+12⇒x3+1x3=64−12=52.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 4^3 = x^3 + \dfrac{1}{x^3} + 3 \times 4 \\[1em] \Rightarrow 64 = x^3 + \dfrac{1}{x^3} + 12 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 64 - 12 = 52.

Hence, x3+1x3=52.x^3 + \dfrac{1}{x^3} = 52.

(ii) Given,

x+1x=4x + \dfrac{1}{x} = 4

Using identity,

⇒(x+1x)2−(x−1x)2=4⇒(4)2−(x−1x)2=4⇒16−(x−1x)2=4⇒(x−1x)2=16−4⇒(x−1x)2=12⇒(x−1x)=12⇒(x−1x)=±23.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (4)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 16 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 16 - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 12 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{12}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.

Hence, (x−1x)=±23.\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.

(iii) Given,

(x−1x)=±23\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}

Case 1:

(x−1x)=23\Big(x - \dfrac{1}{x}\Big) = 2\sqrt{3}

We know that,

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

Substituting values we get :

⇒(x3−1x3)=(23)3+3(23)⇒(x3−1x3)=(8×33)+(63)⇒(x3−1x3)=(243)+(63)⇒(x3−1x3)=303\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (2\sqrt{3})^3 + 3(2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (8 \times 3\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (24\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 30\sqrt{3} \\[1em]

Case 1:

(x−1x)=−23\Big(x - \dfrac{1}{x}\Big) = -2\sqrt{3}

We know that,

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

Substituting values we get :

⇒(x3−1x3)=(−23)3+3(−23)⇒(x3−1x3)=(−8×33)+(−63)⇒(x3−1x3)=(−243)−(63)⇒(x3−1x3)=−303\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-2\sqrt{3})^3 + 3(-2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-8 \times 3\sqrt{3}) + (-6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-24\sqrt{3}) - (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = -30\sqrt{3} \\[1em]

Hence, (x3−1x3)=±303.\Big(x^3 - \dfrac{1}{x^3}\Big) = \pm 30\sqrt{3}.

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