Question 133

If $$\frac{x}{x^2-2x+1}=\frac{1}{3}$$ then the value of $$x^{3}+\frac{1}{x^3}$$ is :

Solution

$$x^{3}+\frac{1}{x^3}$$ = $$(x + \frac{1}{x})^3$$ - 3(x+ $$\frac{1}{x}$$) ....................(1)

Now,

$$\frac{x}{x^2-2x+1}=\frac{1}{3}$$

$$(x^2 -2x+1)=3x$$

$$(x^2-5x+1)=0$$

$$(x^2 + 1)$$ = 5x............(2)

using equation 1 and 2

$$(x + \frac{1}{x})^3$$ - 3(x+ $$\frac{1}{x}$$) = $$(\frac{x^2 + 1}{x})^3$$ - 3$$\frac{(x^2+1)}{x}$$ = 125 - 15 = 110

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