If x^4 - 62 x^2 + 1 = 0, where x > 0, then the value of x^3 + x^{-3} is:

Question 52

If $$x^4 - 62 x^2 + 1 = 0$$, where $$x > 0$$, then the value of $$x^3 + x^{-3}$$ is:

Solution

$$x^4-62x^2+1=0$$

$$x^4+1=62x^2$$

$$x^2+\frac{1}{x^2}=62$$

$$x^2+\frac{1}{x^2}+2=64$$

$$\left(x+\frac{1}{x}\right)^2=64$$

$$x+\frac{1}{x}=8$$.......(1)

$$\left(x+\frac{1}{x}\right)^3=512$$

$$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=512$$

$$x^3+\frac{1}{x^3}+3\left(8\right)=512$$

$$x^3+\frac{1}{x^3}+24=512$$

$$x^3+\frac{1}{x^3}=488$$

Hence, the correct answer is Option C

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