Ominous powers

The determinant of the coefficient matrix of the LHS is non-zero (-6). The column vector of the RHS is all zeroes. To solve for a, b and c, we replace a column in the coefficient matrix with the column vector from the RHS. The determinant of a matrix with any column all zero is also zero. Therefore, a = b = c = 0/(-6) = 0. Zero to any non-zero power is also 0, so the answer is 0.

a+b+c=0, 4a+2b+c=0, a+2b+4c=0 (4a+2b+c)-(a+b+c)=3a+b=0 (a+2b+4c)-(a+b+c)=b+3c=0 (4a+2b+c)-(a+2b+4c)=3a-3c=0 a=c from 3a-3c=0 b=-3a from 3a+b=0 a+a-3a=0 -a=0, a=0 or c+c-3c=0, c=0 so a=0, b=0, c=0

$a+b+c=0\rightarrow Eq.1$
$4a+2b+c=0\rightarrow Eq.2$
$a+2b+4c=0\rightarrow Eq.3$
Subtracting $Eq.3$ and $Eq.2$ , we get,
$3a-3c=0$
$\Rightarrow \boxed {a=c}\rightarrow Eq. 4$
Now, using $Eq.4$ in $Eq.2$ , we get,
$4c+2b+c=0$
$\Rightarrow b=\frac{(-5c)}{2}$
So, $a:b:c=1:\frac{(-5)}{2}:1$
So, $a=x$ , $b=\frac{(-5x)}{2}$ and $c=x$ .
Putting these values in $Eq.1$ ,
$x+\frac{(-5x)}{2}+x=0$
$2x+ \frac{(-5x)}{2}=0$
$\Rightarrow \frac{(-x)}{2}=0$
$\Rightarrow x=0$
So,
$\boxed {a=x=0}$ , $\boxed {b=\frac{(-5x)}{2}=0}$ & $\boxed {c=x=0}$ .
Hence, $\boxed {a^{2013}+b^{2013}+c^{2013}=0}$ .

solve for the variables, a = b = c = 0, so the answer is 0