A.Math problem

a. alpha = A Belta = B
(A + B )^3 - (3A^2B + 3AB^2)
= ( A + B )^3 - 3AB(A + B ) (e題我係用binomial做既,但係都可以用identity做, 不過本人比較鐘意用e個方法)
since A + B = -b
AB = c
so: (-b)^3 - 3(c)(-b)
b: (A + B )^4 - (4A^3B + 6A^2B^2+4AB^3)
(A+B)^4 - 2AB(2A^2 + 2B2 + 3AB)
= (A+B)^4 - 2AB(2(A+B)^2-4AB+3AB)
= (-b)^4 - 2(c)(2(-b)^2-4(c)+3(c))
你想打個d古怪符號, 你去yahoo search 符號機啦

2008-10-01 14:31:30 補充：
3次方你都可以唔用expansion 用identity做. 但係如果你4次方都唔用expansion. 你用兩次2次方會仲煩

2008-10-01 14:33:38 補充：
符號方面其實係window入面已經有d古怪字打. 例如 alt 41400 會出 ☆
ed係乜我自問唔知, 但係我合口1有e樣野存在, 你去word都會搵到個地方可以入ed 符號

Let the 2 roots be m and n. So,
m + n = -b...........(1)
mn = c .................(2)
m^2 + n^2 = (m +n )^2 - 2mn = b^2 - 2c..............(3)
m^2n^2 = c^2................(4)
(a)
(m + n)^3 = m^3 + n^3 + 3m^n + 3mn^2 = m^3 + n^3 + 3mn(m +n ). That is
-b^3 = m^3 + n^3 + 3c(-b)
So m^3 + n^3 = 3cb - b^3.
(b)
From (3) and (4), we know that m^2 and n^2 are the roots of equation
x^2 + ( 2c - b^2)x + c^2 = 0.
or x^2 + Bx + C = 0 where B = (2c - b^2) and C = c^2.
So m^4 + n^4 = (m^2)^2 + (n^2)^2 = B^2 - 2C = (2c - b^2)^2 - 2(c^2).