Answer
$0$
Work Step by Step
Use the rule $a^{mn}=\left(a^m\right)^n$ to obtain:
$i^6+i^4+i^2+1\\
=i^{2\cdot3}+i^{2\cdot2}+i^2+1\\
=(i^2)^3+(i^2)^2+i^2+1$
Use $i^2=-1$.
$=(-1)^3+(-1)^2-1+1$
$=-1+1+0$
$=0$
Hence, the solution in the standard form is $0$.
