Foundations, part 2

Algebra Level pending

If ( 1 + 1 x 2 ) 3 = ( 1 + 1 x 3 ) 2 \left(1+\dfrac{1}{x^2}\right)^3 = \left(1+\dfrac{1}{x^3}\right)^2 for x ≠ 0 x\neq 0 , find x + 1 x x+\dfrac{1}{x} .

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The answer is 0.667.

Ravneet Singh by Ravneet Singh , 30, India

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2 solutions

Chew-Seong Cheong
Mar 21, 2017

( 1 + 1 x 2 ) 3 = ( 1 + 1 x 3 ) 2 ( x 2 + 1 x 2 ) 3 = ( x 3 + 1 x 3 ) 2 ( x 2 + 1 ) 3 = ( x 3 + 1 ) 2 x 6 + 3 x 4 + 3 x 2 + 1 = x 6 + 2 x 3 + 1 3 x 4 − 2 x 3 + 3 x 2 = 0 Since x ≠ 0 3 x 2 − 2 x + 3 = 0 Dividing both sides by x 3 x − 2 + 3 x = 0 ⟹ x + 1 x = 2 3 ≈ 0.667 \begin{aligned} \left(1+\frac 1{x^2}\right)^3 & = \left(1+\frac 1{x^3}\right)^2 \\ \left(\frac {x^2+1}{x^2}\right)^3 & = \left(\frac {x^3+1}{x^3}\right)^2 \\ \left(x^2+1\right)^3 & = \left(x^3+1\right)^2 \\ x^6 + 3x^4 + 3x^2 + 1 & = x^6 + 2x^3+ 1 \\ 3x^4 - 2x^3 + 3x^2 & = 0 & \small \color{#3D99F6} \text{Since }x \ne 0 \\ 3x^2 - 2x + 3 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }x \\ 3x - 2 + \frac 3x & = 0 \\ \implies x + \frac 1x & = \frac 23 \approx \boxed{0.667} \end{aligned}

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Rahil Sehgal
Mar 21, 2017

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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