If $$x + \frac{1}{x} \ne 0$$ and $${x^3} + \frac{1}{{{x^3}}} = 0{\text{,}}$$ then the value $${\left( {x + \frac{1}{x}} \right)^4}$$ is?
A. 9
B. 12
C. 15
D. 16
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {x^3} + \frac{1}{{{x^3}}} = 0{\text{ }} \cr & {\text{It is possible only when}} \cr & x + \frac{1}{x} = \sqrt 3 {\text{ }} \cr & {\text{So, }}{\left( {x + \frac{1}{x}} \right)^4} = {\left( {\sqrt 3 } \right)^4} \cr & \Leftrightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$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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