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SSC CGL 24 Aug 2021 Shift 3 Paper

Question : 53 of 100

Marks: +1, -0

Given that,

x^{8}-34 x^{4}+1=0, x > 0

(x^{3}-x^{-3})

Solution:

Given,

x^{8}-34 x^{4}+1=0

x^{8}+1=34 x^{4}

Divided by

x^{4}

on both sides,

\;x^{4}+\frac{1}{x^{4}}=34

\Rightarrow\;x^{4}+\frac{1}{x^{4}}+2=34+2

\Rightarrow \left(x^{2}+\frac{1}{x^{2}}\right)^{2}\;\;=36

\Rightarrow \left(x^{2}+\frac{1}{x^{2}}\right)=\sqrt{36}=6

\Rightarrow \left(x^{2}+\frac{1}{x^{2}}\right)=6

\; \text{ Now, }\; \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}=6-2=4

\left(x-\frac{1}{x}\right)=\sqrt{4}=2

Take cube on both sides,

\left(x-\frac{1}{x}\right)^{3}=(2)^{3}

\because\;\; x^{3}-\frac{1}{x^{3}}-3 \times x \times \frac{1}{x}\left(x-\frac{1}{x}\right)=8

\;\; [ \because (a-b)^{3}=a^{3}-b^{3}-3 a b(a-b) ]

\Rightarrow \;\; x^{3}-\frac{1}{x^{3}}=8+3 \times 2=8+6

\Rightarrow \;\; x^{3}-\frac{1}{x^{3}}=14

x^{3}-x^{-3}=14

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