If $$x - \frac{3}{x} = 6,\,x \ne 0,$$ then the value of $$\frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}}$$
A. 80
B. 270
C. 54
D. 90
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}} \cr & = \frac{{x\left( {{x^3} - \frac{{27}}{{{x^3}}}} \right)}}{{x\left( {x - 3 - \frac{3}{x}} \right)}}.....\left( {\text{i}} \right) \cr & {x^3} - \frac{{27}}{{{x^3}}} \cr & = {6^3} + 3 \times 3 \times 6 \cr & = 216 + 54 \cr & = 270 \cr & \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - \frac{3}{x} - 3}} = \frac{{270}}{{6 - 3}} = 90 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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