There Are Some Interesting Cases To Consider

There are three options for which the given equation can be satisfied:

• (i) the exponent $x + 1 = 0 \Longrightarrow x = -1,$ for which the base $x^{2} - 3x + 1 = 5 \ne 0$ ;

• (ii) the base $x^{2} - 3x + 1 = 1 \Longrightarrow x(x - 3) = 0 \Longrightarrow x = 0, 3$ ;

• (iii) the base $x^{2} - 3x + 1 = -1 \Longrightarrow x^{2} - 3x + 2 = (x - 1)(x - 2) = 0 \Longrightarrow x = 1,2,$ and the exponent $x + 1$ is even. This is only the case for $x = 1.$

Thus the given equation is satisfied for $x = -1, 0, 1, 3,$ the sum of which is $\boxed{3}.$

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

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

$\left(x + 1 \right) \cdot\ln\left(x^2 - 3 \cdot x + 1 \right) = 0$

$\ln\left(x^2 - 3 \cdot x + 1 \right) = 0$

$\mathrm{e}^{ \ln\left(x^2 \thinspace - \thinspace 3 \cdot x \thinspace + \thinspace 1 \right)} = \mathrm{e}^{0}$

$x^2 - 3 \cdot x + 1 = 1$

$x^2 - 3 \cdot x = 0$

$x\cdot\left(x - 3\right) = 0$

$x = 0 \: | \: x - 3 = 0$

$x = 0 \: | \: x = 3$

$3+0 =\boxed{3}$

If n^k = 1 then n = 1^(1/k). The only real solution is when n = 1. So x^2 - 3x + 1 = 1 or x^2 -3x = 0 so x = x = 0 or x = 3 summing to 3.

I simply factorized the quadratic equation and sum up the factors that were 1+2=3.

(x^2-3x+1)^ (x+1) = 1.

as 1^y =1 for any value of y,

(x^2-3x+1)^(x+1) = 1^(x+1)

=> x^2-3x+1 =1

=> x^2-3x = 0;

=> x = 0,3. :)

Guys I don't understand why we equating x^2 -3x +1 to -1

( x^2-3x+1 )^x+1 = 1 x^2-3x+1 = 1^(1/x+1) becose 1^anything=1 x^2-3x+1 = 1 x^2-3x = 0 x(x-3)=0 x=3,0 now putting the value of x in equation ( 9-9+1)^2= 1^2 =1 @ x=3 (0-0+1)^1= 1^1= 1 @ x=0