Let the minimum value of F (x) = ax ^ 3 + BX ^ 2 + CX be - 8, and the image of its derivative function y = f '(x) passes through the point (- 2,0) (2 / 3,0) My equation is 64x3a-16b + C = 0 -8a+4b-2c=0 (8 / 27) a + (4 / 9) B + (2 / 3) C = 0, how to solve these three equations and find the process B = - C and what happened? Everybody speed... T-T

Let the minimum value of F (x) = ax ^ 3 + BX ^ 2 + CX be - 8, and the image of its derivative function y = f '(x) passes through the point (- 2,0) (2 / 3,0) My equation is 64x3a-16b + C = 0 -8a+4b-2c=0 (8 / 27) a + (4 / 9) B + (2 / 3) C = 0, how to solve these three equations and find the process B = - C and what happened? Everybody speed... T-T

Yuyuxiaofan, you seem to have a wrong understanding of the concept of extremum. Extremum is the value of function, not the value of independent variable. The minimum value of F (x) = ax ^ 3 + BX ^ 2 + CX is - 8, and the image of its derivative function y = f '(x) passes through the point (- 2,0) (2 / 3,0), so f' (x) = 3ax ^ 2 + 2bx + C = 0, and the two roots are - 2,2 / 3, so - 2 + 2 / 3 = - 2b / 3