One eighth (x-1) ^ 5 = A0 + a1x + a2x ^ 2 + +a5x^5 Find the A0 power of 3, A1 power of 27, A2 power of 3, A3 power of 27, A4 power of 3, A5 power of 27 There's 1 / 8 in front of us. That's why we can't count... Depressed~

One eighth (x-1) ^ 5 = A0 + a1x + a2x ^ 2 + +a5x^5 Find the A0 power of 3, A1 power of 27, A2 power of 3, A3 power of 27, A4 power of 3, A5 power of 27 There's 1 / 8 in front of us. That's why we can't count... Depressed~

Expand (1 / 8) (x-1) ^ 5
Get (1 / 8) (x ^ 5-5x ^ 4 + 6x ^ 3-2x ^ 2 + x-1)
So A0 = - 1 / 8, A1 = 1 / 8, A2 = - 1 / 4, A3 = 3 / 4, A4 = - 5 / 8, A5 = 1 / 8
The topic of junior high school is so complicated Isn't it true that there should be no 1 / 8 in front of it, otherwise the final number can't be calculated, only in the form of index