(9х+12/х^3-64)-1/х^2+4х+16=1/х-4

$\frac{9x+12}{ x^{3}-64 } - \frac{1}{ x^{2}+4x+16} = \frac{1}{x-4} \\ \frac{9x+12}{ x^{3}-4 ^{3} } - \frac{1}{ x^{2}+4x+16} = \frac{1}{x-4} \\ \frac{9x+12}{ (x-4)( x^{2} +4x+16)} - \frac{1}{ x^{2}+4x+16} = \frac{1}{x-4} /*(x-4)( x^{2} +4x+16) \neq 0 \\ (x-4)( x^{2} +4x+16) \neq 0 \\ x-4 \neq 0 \\ x \neq 4 \\ x^{2} +4x+16=0 \\ D=16-64=-48 \\ \\ \frac{(x-4)( x^{2} +4x+16)(9x+12)}{(x-4)( x^{2} +4x+16)} - \frac{(x-4)( x^{2} +4x+16)}{( x^{2} +4x+16) } = \frac{(x-4)( x^{2} +4x+16)}{(x-4)} \\$

$9x+12-(x-4)= x^{2} +4x+16\\9x+12-x+4- x^{2} -4x-16=0 \\ - x^{2} +4x=0/*(-1) \\ x^{2} -4x=0 \\ x(x-4) =0 \\ x=0 \\ x-4=0 \\ x=4 \\ x \neq 4$

Ответ: x=0