This is effectively a sixth-degree polynomial, and we can solve it without invoking Fermat's Last Theorem.
All three binomials have x-1 as a root. Then we can write the equation as: (x-1)^3*(x+4)^3 + (x-1)^3*(2x-3)^3 = (x-1)^3*(3x+1)^3
Therefore x-1 is a factor and then x=1 is a root. That leaves a cubic in the form of (x+4)^3 + (2x-3)^3 = (3x+1)^3
Now the next observation is the sum of the two binomials on the left equals the binomial on the right.
So use the sum of cubes formula to write this as (3x+1)*[(x+4)^2 - (x+4)*(2x-3)+ (2x-3)^2] = (3x+1)^3
Therefore 3x+1=0 is a factor and then x=-1/3 is a root.
I can rearrange the prior equation a couple more ways:
(x+4)^3 - (3x+1)^3 = (-2x+3)^3
(2x-3)^3 - (3x+1)^3 = (-x-4)^3
And then using a very similar trick with differences of cubes we get -2x+3 and -x-4 are factors and then x=3/2 and x=-4 are also both roots.
This is three roots of a cubic equation, so those must be all the roots.
The solutions of the given equation are x=1, x=-1/3, x=3/2 and x=-4.
Solution
