Помогите решить уравнение

Решите задачу:

$(x-3) \sqrt[3]{ \frac{x-3}{x+4} } -(x+4) \sqrt[3]{ \frac{x+4}{x-3} } =7\; ,\; ODZ:\; x\ne -4,\; x\ne 3\\\\\\t^3= \frac{x-3}{x+4} \; ,\; x-3=xt^3+4t^3\; ,\; x-xt^3=4t^3+3\; ,\\\\ x(1-t^3)=4t^3+3\; ,\; \; x= \frac{4t^3+3}{1-t^3} \\\\x-3= \frac{4t^3+3}{1-t^3} -3== \frac{4t^3-3-3+3t^3}{1-t^3} = \frac{7t^3}{1-t^3} \; ;\\\\x+4= \frac{4t^3+3}{1-t^3} +4= \frac{4t^3+3+4-4t^3}{1-t^3} = \frac{7}{1-t^3} \; ;\\\\\\ \frac{7t^3}{1-t^3} \cdot t- \frac{7}{1-t^3} \cdot t=7\\\\ \frac{7t^4-7t}{1-t^3}-7=0$

$\frac{7t^4-7t-7+7t^3}{1-t^3} =0\\\\ \frac{7(t^4+t^3-t-1)}{1-t^3} =0\; \; \Rightarrow \; \; \left \{ {{t^4+t^3-t-1=0} \atop {t^3\ne 1}} \right. \; \left \{ {{t^3(t+1)-(t+1)=0} \atop {t\ne 1}} \right. \\\\ \left \{ {{(t+1)(t^3-1)=0} \atop {t\ne 1}} \right. \; \left \{ {{(t+1)(t-1)(t^2+t+1)=0} \atop {t\ne 1}} \right. \; \left \{ {{t_1=-1,\; t_2=1} \atop {t\ne 1}} \right. \; \Rightarrow \; \; t=-1\\\\t=-1\; \; \to \; \; t^3=-1\; \; \to \; \; \frac{x-3}{x+4} =-1\\\\ \frac{x-3}{x+4} +1=0$

$\frac{x-3+x+4}{x+4} =0\\\\ \frac{2x+1}{x+4} =0\\\\2x+1=0\; ,\; x\ne -4\\\\x=-\frac{1}{2}\\\\Otvet:\; \; -\frac{1}{2}\; .$