$\left(x^2-3x+1\right)^{12-x}=1$

Case 1: $a^{0} = 1$ for any $a$ .

$12-x = 0$

Which gives $x = 12$ .

Case 2: $1^{a} = 1$ for any $a$ .

$x^{2} - 3x + 1 = 1$

Which gives $x = 0, 3$ .

Case 3: $(-1)^{a} = 1$ for any even number $a$ . *We have to check that $12 - a$ is even. (or $a$ is even.)

$x^{2} - 3x + 1 = -1$

Which gives $x = 1, 2$ . But $1$ can't be used when we test the equation.

Therefore, the sum = $12+3+0+2 = \boxed{17}$ .

1 raised to any integer is 1.

so ${ x }^{ 2 }-3x+1=1$

${ x }^{ 2 }-3x=0$

$x(x-3)=0$

so $x =3$ and $x=0$

any integer raised to 0 is also 1

so $12-x=0$

$x=12$

$-1$ integer raised to any even integer is 1

so ${ x }^{ 2 }-3x+2=0$

$(x-1)(x-2)$

so $x=1$ and $x=2$

but $12-x$ should be even. so $x=1$ cannot be considered as $12-1=11$

so the sum of all integral values of x is

$3+0+12+2=17$

0,3 and 1,2 are the four solutions after solving ( once for 1 and for -1) .We cannot take 1 because -1 will come .Now if the power is 0 then also the answer will come 1 so 12 is another solution . So answer is 0+3+2+12 = 17.