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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