Question : If $x+\frac{1}{2 x}=3$, then evaluate $8 x^3+\frac{1}{x^2}$.
Option 1: 212
Option 2: 216
Option 3: 180
Option 4: 196
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Correct Answer: 180
Solution :
$x+\frac{1}{2x}=3$
On multiplying 2 on both sides,
$⇒2x+\frac{1}{x}=6$
Now cubing both sides, we get,
$⇒(2x+\frac{1}{x})^{3}=6^{3}$
$⇒8x^{3}+\frac{1}{x^3}+3(4x^2)(\frac{1}{x})+3(2x)(\frac{1}{x^2})=216$
$⇒8x^{3}+\frac{1}{x^3}+12x+\frac{6}{x}=216$
$⇒8x^{3}+\frac{1}{x^3} = 216-6(2x+\frac{1}{x})$
$⇒8x^{3}+\frac{1}{x^3} = 216-6(6)$
$⇒8x^{3}+\frac{1}{x^3} = 216-36$
$⇒8x^{3}+\frac{1}{x^3} = 180$
Hence, the correct answer is 180.
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