If ${log}_{x + 1} {(x^{2} + x - 6)}^{2} = 4$ , then the value(s) of twice the sum of all possible values of $x$ is/are

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Solution

${log}_{x + 1} {(x^{2} + x - 6)}^{2} = 4$
$(x + 1)^{4} = {(x^{2} + x - 6)}^{2}$
$\Rightarrow 2 x^{3} + 17 x^{2} + 16 x - 35 = 0$
$\Rightarrow (x - 1) (2 x^{2} + 19 x + 35) = 0$
$\Rightarrow (x - 1) (2 x + 5) (x + 7) = 0$
$\Rightarrow x = 1, - 7, - \frac{5}{2}$
Now, from the given equation it follows that
$x + 1 > 0, x + 1 \neq 1, x^{2} + x - 6 \neq 0$
$x > - 1, x \neq 0, x \neq - 3, x \neq 2$
$\Rightarrow x \in (- 1, 0) \cup (0, 2) \cup (2, \infty)$
So, $x = 1$ lies in this range.
So, required value $= 2$

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