How do you solve #4 /( x^2 + 3x - 10) - 1 / (x^2 - x -6) = 3 / (x^2 - x -12) #?

Let us first look at the factors of the denominators:

#x^2+3x-10 = (x+5)(x-2)#

#x^2-x-6 = (x-3)(x+2)#

#x^2-x-12 = (x-4)(x+3)#

Note that there are no linear factors in common, so we might as well leave the quadratics unfactorised for now:

Subtract the right hand side of the given equation from the left to get:

#0 = 4/(x^2+3x-10)-1/(x^2-x-6)-3/(x^2-x-12)#

#color(white)(0) = (4(x^2-x-6)(x^2-x-12)-(x^2+3x-10)(x^2-x-12)-3(x^2+3x-10)(x^2-x-6))/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

#color(white)(0) = (4(x^4-2x^3-17x^2+18x+72)-(x^4+2x^3-25x^2-26x+120)-3(x^4+2x^3-19x^2-8x+60))/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

#color(white)(0) = ((4x^4-8x^3-68x^2+72x+288)-(x^4+2x^3-25x^2-26x+120)-(3x^4+6x^3-57x^2-24x+180))/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

#color(white)(0) = (color(red)(cancel(color(black)(4x^4)))-8x^3-68x^2+72x+288-color(red)(cancel(color(black)(x^4)))-2x^3+25x^2+26x-120-color(red)(cancel(color(black)(3x^4)))-6x^3+57x^2+24x-180)/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

#color(white)(0) = (-16x^3+14x^2+122x-12)/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

#color(white)(0) = (-2(8x^3-7x^2-61x+6))/((x^2+3x-10)(x^2-x-6)(x^2-x-12))#

So the solutions of the original equation are the roots of:

#8x^3-7x^2-61x+6 = 0#

To find how to solve this, see: https://socratic.org/s/aydGmACm

The roots are:

#x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))#

for #n = 0, 1, 2#