(1) (x^2+7x+12) = 0 (any number to the 0 power is 1)
(2) (x^2 +3x+1) = 1 (1 raised to any power is 1)
(3) (x^2+3x+1) = -1 AND (x^2+7x+12) is even (-1 raised to an even power is 1)
(1)
x = -3 and x = -4 are solutions.
(2)
x = -3 and x = 0 are solutions (x = -3 is already known to be a solution).
(3)
x = -1 and x = -2 are solutions IF they make x^2+7x+12 = (x+3)(x+4) even.
Note that, for both x = -1 and x = -2, (x+3)(x+4) are consecutive integers. One of them has to be even and one odd; so their product is even.
Therefore, both x = -1 and x = -2 are solutions.
