Question

if a^2 + a + 1=0 then a^9 is?

Answer 1

You could use the quadratic formula to find a then raise it to the 9th power.

Answer 2

I'm not sure if there should be a trick here though or not...

Answer 3

I will try to think of a trick.

Answer 4

I suppose this is a trig question actually.

Answer 5

You can complete the square on the left, solve for a and then raise it to the 9th power.

Answer 6

You could Solve a^2+a+1=0
Then a=y+zi=e^{i theta}
theta is easy to find in this case

Answer 7

y and z are real numbers of course

Answer 8

so you would have a=e^{i (theta}
Raise both sides the the 9th power

Answer 9

that would be easier than doing a^9 algebraically

Answer 10

\[a^2+a+1=0\]

so \[a^9+a^8+a^7=0\Rightarrow a^9=-(a^8+a^7)\]

but \[a^6\times(a^2+a+1=0)\Rightarrow a^8+a^7=-a^6\]

thus
\[a^9=-(a^8+a^7)=-(-a^6)=a^6\]

...keep doing this.

Answer 11

it was asked in entrance exam . and had 1 mins to solve

Answer 12

Well I suggested a trig way. Do you know trig?

Answer 13

it only takes about a min ;)

Answer 14

That is an interesting way zarkon

Answer 15

there is a pattern ...

\[a^9=a^6=a^3=a^0=1\]

Answer 16

a^9=-1

Answer 17

no...it is 1

Answer 18

to verify I just solved it on my calculator and raised the answer to the 9th power...got 1

Answer 19

oops I had 1/2+- sqrt(3)/2

Answer 20

forgot to put - in front of the 1/2

Answer 21

\[a=\frac{-1}{2} \pm \frac{\sqrt{3}}{2} \\ a=e^{\frac{2\pi}{3} i} \\ a^9=e^{6\pi i}\]

Answer 22

=1

Answer 23

sorry my bad stupid -

Answer 24

that seems like a lot of work for this problem :)

Answer 25

That isn't a lot of work if you know the unit circle

Answer 26

But that is my opinion.

Answer 27

thanks all
:)

Answer 28

a^2 + a + 1=0

(a-1) (a^2+a+1) =0

a^3 -1 = 0
a^3 =1
a=1
a^9 =1

took about 4 secs

Answer 29

you can't say \(a=1\)
(because it is not...that zero was introduced when you multiplied by a-1)

but you can go from \(a^3=1\) to \((a^3)^3=(1)^3\Rightarrow a^9=1\)

Answer 30

right!
sorry, my bad

Answer 31

nice idea though.