If a2 + a + 1 = 0, then the value of a9 is?
A. 2
B. 3
C. 1
D. 0
Answer: Option C
Solution (By Examveda Team)
$${a^2} + a + 1 = 0$$\[\left[ \begin{array}{l} {a^3} + {1^3} = \left( {a + 1} \right)\left( {{a^2} + a + 1} \right)\\ {a^3} - {1^3} = \left( {a - 1} \right)\left( {{a^2} + a + 1} \right) \end{array} \right]\]
$$\eqalign{ & \therefore \left( {{a^3} - 1} \right) = \left( {a - 1} \right) \times 0 \cr & \Rightarrow {a^3} - 1 = 0 \cr & \Rightarrow {a^3} = 1 \cr & \Rightarrow {\left( {{a^3}} \right)^3} = {1^3} \cr & \Rightarrow {a^9} = 1 \cr} $$
This Question Belongs to Arithmetic Ability >> Algebra
If we use the value-(1) for ‘a’ in the question, the result stands
1^2+1+1=0
or
3=0
How is it possible?
wrong formula
a3+b3 = (a+b ) a2-ab+b2
Actually ( a^3 + 1^3)=(a+1)(a^2+a+1)