If $${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$$ then find the value of $${a^{30}}$$ + $${a^{24}}$$ + $${a^{18}}$$ + $${a^{12}}$$ + $${a^6}$$ + $$1$$ = ?
A. 0
B. -27
C. 1
D. -1
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Cube both sides}} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} + 1 = 0 \cr & \therefore {\text{ }}{a^{30}} + {a^{24}} + {a^{18}} + {a^{12}} + {a^6} + 1 = ? \cr & \Rightarrow {\text{ }}{a^{24}}\left( {{a^6} + 1} \right) + {a^{12}}\left( {{a^6} + 1} \right) + {a^6} + 1 \cr & \Rightarrow {a^{24}}\left( 0 \right) + {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$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. $$\frac{8}{6}$$
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