Question

can anyone plz help with this question... If (a^a) + 1 = a, then the value of (a^12) + (a^6) + 1 is?

(a) 2 (b) 3 (c) -3 (d) 1

Answer 1

are you really sure it's `a^a + 1 = a` and not `a^2 + 1 = a` ?

Answer 2

maybe some typing mistake.. i have found this question as it is on a website. there they have written a^a.
Well if it was a^2 then what would it be?

Answer 3

Here is one dumb way to work it if you know some complex numbers :

\(a^2 -a+1 = 0 \implies a = e^{\pm i \pi/3}\)

so,

\(a^{12} + a^6 + 1 \\~\\= e^{\pm i \pi/3*12}+ e^{\pm i\pi/3*6}+1\\~\\=e^{\pm i4\pi}+e^{\pm i2\pi}+1\\~\\ = 1+1+1\\~\\=3\)

Answer 4

pretty sure there is a clever way to work it w/o complex numbers..

Answer 5

\[a^2+1=a\Rightarrow a^2=a-1\]

so

\[a^2-a=-1\Rightarrow a(a-1)=-1\]

then

\[a^{12}+a^6+1=a^4a^8+a^2a^4+1=a^4(a^2)^4+a^2(a^2)^2+1\]


\[=a^4(a-1)^4+a^2(a-1)^2+1\]

\[=[a(a-1)]^4+[a(a-1)]^2+1=(-1)^4+(-1)^2+1=1+1+1=3\]

Answer 6

thanks to all :-)