Challenging Pure Maths - LQ

a_0x^3+3a_1x^2+3a_2x+a_3=0
x^3+3(a_1/a_0)x^2+3(a_2/a_0)x+(a_3/a_0)=0
Using x=y+w, it can be shown that when w=-a_1/a_0, the formula will change to y^3+3H_1y+G_1=0
where 3H_1=3(-a_1^2/a_0+a_2) and G_1=2(a_1^3/a_0^2)-3(a_1a_2/a_0^2)+a_3
Now let z=a_0y, the equation becomes
z^3+3Hz+G=0 where
H=-a_1^2+a_0a_2,G=2a_1^3-3a_0a_1a_2+a_0^2a_3
(ii) p+q+r=0,pq+pr+qr=3H,pqr=-G
G^2=(pqr)^2
H^3=[(pq+pr+qr)/3]^3=(1/27)(pq+pr+qr)^3
-27(G^2+4H^3)=-27(pqr)^2-4(pq+pr+qr)^3...(1)
Consider this is a polynomial of p
Substitute p=q,r=-2q,then (1) becomes 0, similarly for p=r
So we have (p-q)(p-r) as the factor of (1)
Consider this is a polynomial of q and substitute q=p and q=r
we have zero. So (q-p) and (q-r) is a factor of (1)
Similar for the view of r, so (q-r)^2(r-p)^2(p-q)^2=-27(G^2+4H^3)
x=y+w=z/a_0+w. This means α=p/a_0-(a_1/a_0),β=q/a_0-(a_1/a_0) γ=r/a_0-(a_1/a_0)
And thus (q-r)^2(r-p)^2(p-q)^2=a_0^6(α- β)^2(β-γ)^2(γ-α)^2
a_0^6(α- β)^2(β-γ)^2(γ-α)^2=-27(G^2+4H^3)
D=(1/a_0^2)(-27)(G^2+4H^3) After expansion, we get the result.
(b)(1) D=0, -27(G^2+4H^3)=0=>(q-r)^2(r-p)^2(p-q)^2=0 So at least two roots are equal.
(2) D>0, (q-r)^2(r-p)^2(p-q)^2>0=>p,q,r are not equal
(3) D<0, (q-r)^2(r-p)^2(p-q)^2<0=>one real roots and two conjugate unreal roots.
(iii)(a) x=u+v,
x^3-15x
=(u+v)^3-15(u+v)
=u^3+v^3+3uv(u+v)-15(u+v)
=u^3+v^3+3(uv-5)(u+v)
(b) y=5/x=>x^3+(5/x)^3-126=0
x^6+125-126x^3=0
z^2-126z+125=0
(z-125)(z-1)=0
z=1 or 125
x=1 or 5
y=5 or 1
(c) x^3-15x-126=0
u^3+v^3+3(uv-5)(u+v)=126
u^3+v^3-126=0
3(uv-5)(u+v)=0
=> u=1,v=5
So x=6 is one root
Now x^3-15x-126=(x-6)(x^2+6x+21) and we can find the other two roots
So



2009-06-14 20:56:38 補充：
(b) Using the result that if a complex number is a root of the polynomial, then its conjugate should also be a root

個人意見:回答問題的方式或層次,是以提問人的程度與題意為依據.
比方: http://tw.knowledge.yahoo.com/question/question?qid=1608102301304
或:http://hk.knowledge.yahoo.com/question/question?qid=7009021200378
指出key point處就屬上選,沒必要將dirty work的部份詳細寫出,看懂題
目的人,自然有相當程度,尊重一下他們的智慧吧!
看看身邊的數學相關教科書(論文就更甭提了),是否都很detail呢?

2009-06-18 11:34:25 補充：
本題若我來寫,說不定更簡略,可能又得挨刮了! 哈哈哈!
每個人都有自己的想法,參考參考即可,
所謂最佳解答是提問人來判斷的,提問者接受的作法即屬最佳!

等我睇下我睇唔睇得明先...

2009-06-17 16:58:11 補充：
整條題目我也重新由頭到尾自己做左一次了,咁就等我講番句公道說話LA...

對於#001提到還有四個complex root的問題,如果就咁問x³ = 1和x³ = 125的解,當然係要出齊6個解,但係因為計x都係為左解c part,咁其實不寫complex root也是可以接受的。

而#002的說法亦不無道理,無錯,呢個寫法雖然唔係咁好,但都係廣為人接受的。

2009-06-17 16:58:18 補充：
至於用factor theorem去證明(ii) part a對我來說亦係一個幾新鮮的做法。

而#008提到當中有些step miss左,我都覺得係應該可以寫詳細d既,因為當中應該涉及leading coefficient的問題吧...

而(ii)(b)(1)(2)(3)部的問題,其實都係幾obvious的result,咁我都覺(3)可以再酌量加多dd解釋...不過整體都冇乜問題,因為真係冇乜step可言,focus都係在於文字上的表達吧...

myisland8132，x³ = 1和x³ = 125這兩條方程除了有x = 1和x = 5這兩個解之外，還有那四個complex root呢？

緊記，你正在做的是高考題目，不是會考題目，仲當complex root不是root，冇得原諒！

2009-06-15 01:41:12 補充：
仲有，你寫：
x=1 or 5

y=5 or 1

這樣的寫法是錯的，因為其意思就會變成：
x = 1 , y = 5 and x = 1 , y = 1 and x = 5 , y = 5 and x = 5 , y = 1

2009-06-16 20:19:25 補充：
myisland8132不回應，真是非常討厭！

2009-06-16 20:24:03 補充：
其實這題的(ii)部正是取材自http://db.math.ust.hk/resource_sharing/algebra/sc_d3.pdf。

2009-06-17 05:49:49 補充：
myisland8132的做法還有不少地方有問題：

在(ii)(a)部：
點解代p = q , r = - 2q而得出= 0就會導出(p - q)(p - r) is a factor？肯定冇人會睇得明，因為在pure maths教factorization of cyclic symmetric expressions的時候只係教咗每次換一個variable嘅做法而冇教每次同時換兩個variable嘅做法。

2009-06-17 05:57:08 補充：
check到(q - r)^2(r - p)^2(p - q)^2 is a factor of - 27(G^2 + 4H^3)當然不可以直接寫(q - r)^2(r - p)^2(p - q)^2 = - 27(G^2 + 4H^3)，即是說，中間漏了一些步驟。

在(ii)(b)(1)(2)(3)部：
簡直可以話照抄題目，完全冇做過，就算係obvious都要講聲啦。(1)和(2)話obvious都仲情有可原，但(3)就一定唔可以話obvious。

2009-06-17 06:03:19 補充：
如果冇新回答嘅話，睇嚟myisland8132留下來的蘇州屎都係由我執嘅啦！STEVIE-G™，延長發問時間吧！