dont understand how to do this

$$\begin{array}{rlll} \frac{3}{x^3-12x^2+48x-64}+\frac{111}{64-x^3}&=&0\qquad&(1)\\\\ \frac{3}{x^3-3(4x^2)+3(4^2x)-4^3}+\frac{111}{-(x^3-64)}&=&0\qquad&(2)\\\\ \frac{3}{(x-4)^3}-\frac{111}{x^3-64}&=&0\qquad&(3)\\\\ \frac{3}{(x-4)^3}-\frac{111}{(x-4)(x^2+4x+16)}&=&0\qquad&(4)\\\\ \frac{3(x^2+4x+16)}{(x-4)^3(x^2+4x+16)}-\frac{111(x-4)^2}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(5)\\\\ \frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111(x^2-8x+16)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(6)\\\\ \frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111x^2-8*111x+16*111}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(7)\\\\ \frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}+\frac{-111x^2+888x-1776}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(8)\\\\ \frac{-108x^2+900x-1728}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(9)\\\\ \frac{-36(3x^2-25x+48)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(10)\\\\ x&=&\frac{25\pm 7}{6}\qquad \qquad&(11)\\\\ \end{array}$$

(11) came from the quadratic formula

NOW which line do you need explained?

(4-x)=-1(-4+x)=-(x-4)

Is that your only problem here?

no i dont understand it thats exactly what i tried to do

Thank you melody I understand everything now