Challenging Pure Maths - SQ

1. (β - γ)^2 = (β^2 + γ^2) - 2(βγ)
= [(α^2 + β^2 + γ^2) - α^2] - 2[(αβ+βγ+αγ) - α(α+β+γ - α)]
= [0^2 - 2q - α^2] - 2[q - α(0 - α)]
= 2q - α^2 - 2q - 2α^2
= -3α^2

Therefore the three roots of the new equation are -3α^2, -3β^2, -3γ^2.

Now compute
-3α^2 - 3β^2 - 3γ^2 = -3 [(α+β+γ)^2 - 2(αβ+βγ+αγ)]
= -3[0^2 - 2q] = 6q
(-3α^2)(-3β^2) + (-3β^2)(-3γ^2) + (-3α^2)(-3γ^2)
= 9 [(αβ+βγ+αγ)^2 - 2(αβγ)^2 (α^2 + β^2 + γ^2)]
= 9 [q^2 - 2(-r)^2 (0^2 - 2q)]
= 9 (q^2 - 4qr^2)
(-3β^2)(-3β^2)(-3γ^2) = -27(-r)^2 = -27r^2

Hence, the new equation is
x^3 - 6q x^2 + 9(q^2 - 4qr^2) x + 27r^2 = 0

2. α^2 - βγ = α^2 - [(αβ+βγ+αγ) - α(α+β+γ - α)]
= α^2 - [q - α(p - α)]
= pα - q

Therefore the three roots are pα-q, pβ-q, pγ-q.

Now compute
(pα - q) + (pβ - q) + (pγ - q) = p(α+β+γ) - 3q = - p^2 - 3q
(pα - q)(pβ - q) + (pβ - q)(pγ - q) + (pα - q)(pγ - q)
= p^2 [αβ+βγ+αγ] - 2pq [α+β+γ] + 3q^2
= p^2 q + 2 p^2 q + 3 q^2
= 3q [p^2 + q]
(pα - q)(pβ - q)(pγ - q) = p^3[αβγ] - p^2 q [αβ+βγ+αγ] + p q^2 [α+β+γ] - q^3
= - p^3 r - p^2 q^2 - p^2 q^2 - q^3
= - [p^3 r + 2 p^2 q^2 + q^3]

Hence, the equation is
x^3 + (p^2 + 3q) x^2 + 3q(p^2 + q) x + (p^3 r + 2 p^2 q^2 + q^3) = 0


2009-06-17 19:09:45 補充：
剛剛才發覺原來第二條完全答非所問……OTZ
還以為兩條數是做同樣的步驟添……=.="

2009-06-17 23:35:12 補充：
方法是沒錯，但第二題要做的是express...in terms of p,q,r，
呢樣野我係做第二題中間都有做到，
但我以為呢條同第一條一樣係求equation……但多此一舉。

噢...我只是看了你的方法,未詳細看你的計算過程唷>