\[x+\frac{1}{2x}=2,\]then find the value of \[8{{x}^{2}}+\frac{1}{{{x}^{3}}}.\] [SSC (10+2) 2013] |
A) 48
B) 88
C) 40
D) 44
Correct Answer: C
Solution :
Given, \[x+\frac{1}{2x}=2\] |
On multiplying by 2 both sides, we get |
\[2x+\frac{2}{2x}=4\]\[\Rightarrow \]\[2x+\frac{1}{x}=4\] |
On cubing both sides, we get |
\[{{\left( 2x+\frac{1}{x} \right)}^{3}}={{4}^{3}}\] |
\[\Rightarrow \]\[{{(2x)}^{3}}+{{\left( \frac{1}{x} \right)}^{3}}+3\times 2x\times \frac{1}{x}\left( 2x+\frac{1}{x} \right)=64\] |
\[[\because {{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\,\,(a+b)]\] |
\[\Rightarrow \]\[8{{x}^{2}}+\frac{1}{{{x}^{3}}}+3\times 2x\times \frac{1}{x}\left( 2x+\frac{1}{x} \right)=64\] |
\[\Rightarrow \]\[8{{x}^{3}}+\frac{1}{{{x}^{3}}}+6\times 4=64\] |
\[\Rightarrow \]\[8{{x}^{3}}+\frac{1}{{{x}^{3}}}=64-24=40\] |
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