Question : If $6x^{2}-12x+1=0$, then the value of $27x^{3}+\frac{1}{8x^{3}}$ is:
Option 1: 162
Option 2: 189
Option 3: 207
Option 4: 225
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Correct Answer: 189
Solution :
Given: $6x^{^2}-12x+1=0$
Dividing by $2x$, we get $3x+\frac{1}{2x}=6$
Cubing both sides, $\left (3x \right)^{3}+\left (\frac{1}{2x} \right )^{3}+3\left (3x \right)\left (\frac{1}{2x}\right )\left (3x+\frac{1}{2x} \right )=6^{3}$
⇒ $27x^{3}+\frac{1}{8x^{3}}+3\left (3\right )\left (\frac{1}{2} \right )\left (6\right )=6^{3}$
⇒ $27x^{3}+\frac{1}{8x^{3}}=6^{3}-3\left (3\right)\left (\frac{1}{2}\right)\left (6\right)=189$
Hence, the correct answer is 189.
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