The sum of all distinct roots of the equation $(x^{2} - 11 x + 29)^{(x^{2} - 8 x + 7)} = 1$ is

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Solution

$(x^{2} - 11 x + 29)^{(x^{2} - 8 x + 7)} = 1$
Let $a = x^{2} - 11 x + 29$ and $b = x^{2} - 8 x + 7$

Case $1 :$ $a^{b} = 1$ when $a = 1$
$x^{2} - 11 x + 29 = 1$
$\Rightarrow x^{2} - 11 x + 28 = 0 \Rightarrow x = 4, 7$

Case $2 :$ $a^{b} = 1$ when $b = 0$
$x^{2} - 8 x + 7 = 0 \Rightarrow x = 1, 7$

Case $3 :$ $a^{b} = 1$ when $a = - 1$ and $b$ is an even number.
$x^{2} - 11 x + 29 = - 1$
$\Rightarrow x^{2} - 11 x + 30 = 0 \Rightarrow x = 5, 6$
For $x = 5, LHS = (- 1)^{- 8} = 1 = RHS$
For $x = 6, LHS = (- 1)^{- 5} = - 1 \neq RHS$

Therefore, distinct roots are $1, 4, 5, 7$
Sum of distinct roots $= 4 + 7 + 1 + 5 = 17$

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