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SSC CGL 24 Aug 2021 Shift 3 Paper

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Question : 52 of 100

Marks: +1, -0

If

x^{4}-62 x^{2}+1=0

x>0

, then the value of

x^{3}+x^{-3}

is

500
512
488
364

Solution:

Given,

x^{4}-62 x^{2}+1=0

x^{4}+1=62 x^{2}

Divided by

x^{2}

, we get

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

We know that,

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

\left(x+\frac{1}{x}\right)^{2}=62+2=64

\left[\because (a+b)^{2}=a^{2}+b^{2}+2ab\right]

\left(x+\frac{1}{x}\right)=\sqrt{64}=8 \Rightarrow \left(x+\frac{1}{x}\right)=8

Taking cube on both sides,

\Rightarrow \left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3 \times x

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

\left[\because (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)\right]

\Rightarrow x^{3}+\frac{1}{x^{3}}=512-3 \times 8=512-24

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

or

x^{3}+x^{-3}=488

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