** **The sum of all real values of *x* satisfying the equation \[{\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1\] is

Given, \[(x^{2} - 5x + 5)^{x^{2} + 4x - 60} = 1\]

Clearly, this is possible when

I. \[x^{2} + 4x - 60 = 0\] and \[x^{2} - 5x + 5 \neq 0\]

or

II. \[x^{2} - 5x + 5 = 1\]

III. \[x^{2} - 5x + 5 = - 1\] and \[x^{2} + 4x - 60\] = Even integer.

Case I When \[x^{2} + 4x - 60 = 0\]

⇒ \[x^{2} + 10x - 6x - 60 = 0\]

⇒ \[x(x + 10) - 6(x + 10) = 0\]

⇒ (x + 10)(x - 6) = 0

⇒ x = -10 or x = 6

Note that, for these two values of \[x, x^{2} - 5x + 5 \neq 0\]

Case II When \[x^{2} - 5x + 5 = 1\]

⇒ \[x^{2} - 5x + 4 = 0\]

⇒ \[x^{2} - 4x - x + 4 = 0\]

⇒ x(x - 4) -1(x - 4) = 0

⇒ (x - 4)(x - 1) = 0 ⇒ x = 4 or x = 1

Case III When \[x^{2} - 5x + 5 = -1\]

⇒ \[x^{2} - 5x + 6 = 0\]

⇒ \[x^{2} - 2x - 3x + 6 = 0\]

⇒ x(x - 2) - 3(x - 2) = 0

⇒ (x - 2)(x - 3) = 0

⇒ x = 2 or x = 3.

Now, when x = 2, \[x^{2} + 4x - 60 = 4 + 8 - 60 = - 48\], which is an even integer.

When x = 3, \[x^{2} + 4x - 60 = 9 + 12 - 60 = -39\], which is not an even integer.

Hence the sum of all real values of x = -10 + 6 + 4 + 1 + 2 = 3.