Question : If $x-\frac{3}{x}=6, x \neq 0$, then the value of $\frac{x^4-\frac{27}{x^2}}{x^2-3 x-3}$ is:
Option 1: 80
Option 2: 270
Option 3: 54
Option 4: 90
Correct Answer: 90
Solution :
$x-\frac{3}{x}=6$
⇒ $x^2 - 3 = 6x$
⇒ $x^2 - 6x -3 =0$
⇒ $x^2 - 3x -3 =3x$ ---(1)
($x-\frac{3}{x})^3= 6^3$
⇒ $x^3 - (\frac{3}{x})^3 - 3(x-\frac{3}{x}) = 216$
⇒ $x^3 - (\frac{3}{x})^3 - 9(6) = 216$
⇒ $x^3 - (\frac{3}{x})^3 - 54 = 216$
⇒ $x^3 - (\frac{3}{x})^3 = 270$ --- (2)
Now, $\frac{x^4-\frac{27}{x^2}}{x^2-3 x-3}$
= $\frac{x^4-\frac{27}{x^2}}{3x}$ [from equation (1)]
= $\frac{x^3-\frac{27}{x^3}}{3}$
= $\frac{270}{3}$ [From the above equation (2)]
= 90
Hence the correct answer is 90.
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