$$\begin{array}{rlll}
\frac{3}{x^3-12x^2+48x-64}+\frac{111}{64-x^3}&=&0\qquad&(1)\\\\
\frac{3}{x^3-3(4x^2)+3(4^2x)-4^3}+\frac{111}{-(x^3-64)}&=&0\qquad&(2)\\\\
\frac{3}{(x-4)^3}-\frac{111}{x^3-64}&=&0\qquad&(3)\\\\
\frac{3}{(x-4)^3}-\frac{111}{(x-4)(x^2+4x+16)}&=&0\qquad&(4)\\\\
\frac{3(x^2+4x+16)}{(x-4)^3(x^2+4x+16)}-\frac{111(x-4)^2}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(5)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111(x^2-8x+16)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(6)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111x^2-8*111x+16*111}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(7)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}+\frac{-111x^2+888x-1776}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(8)\\\\
\frac{-108x^2+900x-1728}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(9)\\\\
\frac{-36(3x^2-25x+48)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(10)\\\\
x&=&\frac{25\pm 7}{6}\qquad \qquad&(11)\\\\
\end{array}$$
(11) came from the quadratic formula
NOW which line do you need explained?
$$\begin{array}{rlll}
\frac{3}{x^3-12x^2+48x-64}+\frac{111}{64-x^3}&=&0\qquad&(1)\\\\
\frac{3}{x^3-3(4x^2)+3(4^2x)-4^3}+\frac{111}{-(x^3-64)}&=&0\qquad&(2)\\\\
\frac{3}{(x-4)^3}-\frac{111}{x^3-64}&=&0\qquad&(3)\\\\
\frac{3}{(x-4)^3}-\frac{111}{(x-4)(x^2+4x+16)}&=&0\qquad&(4)\\\\
\frac{3(x^2+4x+16)}{(x-4)^3(x^2+4x+16)}-\frac{111(x-4)^2}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(5)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111(x^2-8x+16)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(6)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111x^2-8*111x+16*111}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(7)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}+\frac{-111x^2+888x-1776}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(8)\\\\
\frac{-108x^2+900x-1728}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(9)\\\\
\frac{-36(3x^2-25x+48)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(10)\\\\
x&=&\frac{25\pm 7}{6}\qquad \qquad&(11)\\\\
\end{array}$$
(11) came from the quadratic formula
NOW which line do you need explained?