If $${\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3{\text{,}}$$ then the value of a18 + a12 + a6 + 1 is?
A. 3
B. 1
C. 0
D. 2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} = - 1 \cr & \Rightarrow \left( {{a^6} + 1} \right) = 0 \cr & {\text{So,}}{a^{18}} + {\text{ }}{a^{12}} + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( {{a^6} + 1} \right) + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$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. $$\frac{8}{6}$$
Join The Discussion