Question : If $2x + {\frac{2}{9x}} = 4$, then the value of $27x^3 + \frac{1}{27x^{3}}$ is:
Option 1: 180
Option 2: 198
Option 3: 234
Option 4: 252
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Correct Answer: 198
Solution :
Given: $2x + {\frac{2}{9x}}= 4$--------------------------(i)
In order to bring 3 as coefficient of $x$, multiply (i) with $\frac{3}{2}$,
$\frac{3}{2}×(2x+{\frac{2}{9x}}) = \frac{3}{2}×4$
⇒$\frac{3}{2}×2x+\frac{3}{2}×\frac{2}{9x} = 6$
⇒ $3x+\frac{3}{9x} = 6$
⇒ $3x+\frac{1}{3x} = 6$-------------------------(ii)
Taking a cube on both sides of equation (ii),
$(3x+\frac{1}{3x})^{3} = 6^{3}$
Since $(a+b)^{3}= a^{3}+b^{3}+3ab(a+b)$.
⇒ $(3x)^{3}+(\frac{1}{3x})^{3}+3(3x)(\frac{1}{3x})(3x+\frac{1}{3x})=6^{3}$
⇒ $27x^{3}+ \frac{1}{27x^{3}}+3×(3x +\frac{1}{3x})=216$
⇒ $27x^{3}+ \frac{1}{27x^{3}}=216–3×(3x+\frac{1}{3x})= 216 – (3×6) =198$
Hence, the correct answer is 198.
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