Решите уравнение: б) (2x+1)(3x^2+1)(4x+1)=0 в)(x-8)(x^2-7x-8)=x^3-8x^2

a) $(2x+1)(3x^2+1)(4x+1)=0\\ 2x+1=0 \\ 2x=-1\\ x=-\frac{1}{2}\\ 3x^2+1=0\\ 3x^2=-1\\ x^2=-\frac{1}{3}\\ x=+-\sqrt{\frac{1}{3}}\\ 4x+1=0\\ 4x=-1\\ x=-\frac{1}{4}\\$

б) $(x-8)(x^2-7x-8)=x^3-8x^2\\ (x-8)(x+1)(x-8)=x^2(x-8)\\ (x-8)^2(x+1)-x^2(x-8)=0\\ (x-8)((x-8)(x+1)-x^2)=0\\ (x-8)(x^2-7x-8-x^2)=0\\ (x-8)(-7x-8)=0\\ x-8=0\\ x=8\\ -7x-8=0\\ -7x=8\\ x=-1\frac{1}{8}$