分式方程 救命！！！

Solve (x² + 2x + 1)/(x² + 5x + 4) + (x – 1)/(x² – 3x – 4) = 0.

(x² + 2x + 1)/(x² + 5x + 4) + (x – 1)/(x² – 3x – 4) = 0
(x + 1)²/((x + 4)(x + 1)) + (x – 1)/((x – 4)(x + 1)) = 0
((x + 1)²(x – 4) + (x – 1)(x + 4))/((x + 4)(x – 4)(x + 1)) = 0
(x + 1)²(x – 4) + (x – 1)(x + 4) = 0
(x² + 2x + 1)(x – 4) + x² + 3x – 4 = 0
x³ + 2x² + x – 4x² – 8x – 4 + x² + 3x – 4 = 0
x³ – x² – 4x – 8 = 0

Let f(x) = x³ – x² – 4x – 8
f(1) = 1³ – 1² – 4 × 1 – 8 = – 12 ≠ 0
∴x – 1 is not a factor.
f(– 1) = (– 1)³ – (– 1)² – 4(– 1) – 8 = – 6 ≠ 0
∴x + 1 is not a factor.
f(2) = 2³ – 2² – 4 × 2 – 8 = – 12 ≠ 0
∴x – 2 is not a factor.
f(– 2) = (– 2)³ – (– 2)² – 4(– 2) – 8 = – 12 ≠ 0
∴x + 2 is not a factor.
f(4) = 4³ – 4² – 4 × 4 – 8 = 24 ≠ 0
∴x – 4 is not a factor.
f(– 4) = (– 4)³ – (– 4)² – 4(– 4) – 8 = – 72 ≠ 0
∴x + 4 is not a factor.
f(8) = 8³ – 8² – 4 × 8 – 8 = 408 ≠ 0
∴x – 8 is not a factor.
f(– 8) = (– 8)³ – (– 8)² – 4(– 8) – 8 = – 552 ≠ 0
∴x + 8 is not a factor.

今次你真係唔想喊救命都唔得啦！因為x³ – x² – 4x – 8冇得factorize！

不過，唔洗驚！呢條equation唔係solve唔倒，只不過要用到三次公式（http://zh.wikipedia.org/wiki/%E4%B8%89%E6%AC%A1%E6%96%B9%E7%A8%8B）之嘛！

Let x = y – h,
Then (y – h)³ – (y – h)² – 4(y – h) – 8 = 0
The term in y² will vanish if – 3h – 1 = 0
Choosing h = – 1/3, the equation becomes
(y + 1/3)³ – (y + 1/3)² – 4(y + 1/3) – 8 = 0
y³ + (3/3)y² + (3/9)y + 1/27 – (y² + (2/3)y + 1/9) – 4y – 4/3 – 8 = 0
y³ + y² + (1/3)y + 1/27 – y² – (2/3)y – 1/9 – 4y – 4/3 – 8 = 0
y³ – (13/3)y – 254/27 = 0, which is a cubic equation which is the depressed cubic.

Let y = u + v,
Then u³ + v³ = 254/27 and uv = 13/9
Hence u³ + (13/(9u))³ = 254/27
u³ + 2197/(729u³) = 254/27
729u^6 + 2197 = 6858u³
729u^6 – 6858u³ + 2197 = 0
u³ = [– (– 6858) ± √((– 6858)² – 4(729)(2197))]/(2 × 729)
u = ³√((6858 ± √40625712)/1458)
　= ³√((6858 ± 972√43)/1458)
　= ³√((127 ± 18√43)/27)
　= (³√(127 ± 18√43))/3

When u = (³√(127 + 18√43))/3,
v = 13/[9[(³√(127 + 18√43))/3]]
　= 13/[3(³√(127 + 18√43))]
　= ³√[2197/(27(127 + 18√43))]
　= ³√[(2197(127 – 18√43))/(27(127 + 18√43)(127 – 18√43))]
　= ³√[(2197(127 – 18√43))/(27(2197))
　= (³√(127 – 18√43))/3

When u = (³√(127 – 18√43))/3,
v = 13/[9[(³√(127 – 18√43))/3]]
　= 13/[3(³√(127 – 18√43))]
　= ³√[2197/(27(127 – 18√43))]
　= ³√[(2197(127 + 18√43))/(27(127 – 18√43)(127 + 18√43))]
　= ³√[(2197(127 + 18√43))/(27(2197))
　= (³√(127 + 18√43))/3

由此可見，u和v是共軛，所以我哋只需選擇其中一組u, v的解即可。

Hence y = u + v or ωu + ω²v or ω²u + ωv,
where ω = e^(2iπ/3) = cos (2π/3) + i sin (2π/3) = – 1/2 + ((√3)/2)i

∴y = (³√(127 + 18√43))/3 + (³√(127 – 18√43))/3 or (– 1/2 + ((√3)/2)i)((³√(127 + 18√43))/3) + (– 1/2 + ((√3)/2)i)²((³√(127 – 18√43))/3) or (– 1/2 + ((√3)/2)i)²((³√(127 + 18√43))/3) + (– 1/2 + ((√3)/2)i)((³√(127 – 18√43))/3)
= (³√(127 + 18√43) + ³√(127 – 18√43))/3 or (– 1/2 + ((√3)/2)i)((³√(127 + 18√43))/3) + (1/4 + ((√3)/2)i – 3/4)((³√(127 – 18√43))/3) or (1/4 + ((√3)/2)i – 3/4)((³√(127 + 18√43))/3) + (– 1/2 + ((√3)/2)i)((³√(127 – 18√43))/3)
= (³√(127 + 18√43) + ³√(127 – 18√43))/3 or (– 1/2 + ((√3)/2)i)((³√(127 + 18√43))/3) + (– 1/2 – ((√3)/2)i)((³√(127 – 18√43))/3) or (– 1/2 – ((√3)/2)i)((³√(127 + 18√43))/3) + (– 1/2 + ((√3)/2)i)((³√(127 – 18√43))/3)
= (³√(127 + 18√43) + ³√(127 – 18√43))/3 or [[(√3) ³√(127 + 18√43) – (√3) ³√(127 – 18√43)]i – ³√(127 + 18√43) – ³√(127 – 18√43)]/6 or [[(√3) ³√(127 – 18√43) – (√3) ³√(127 + 18√43)]i – ³√(127 + 18√43) – ³√(127 – 18√43)]/6

Hence x = y + 1/3
= (1 + ³√(127 + 18√43) + ³√(127 – 18√43))/3 or [(³√(381√3 + 54√129) – ³√(381√3 – 54√129))i + 2 – ³√(127 + 18√43) – ³√(127 – 18√43)]/6 or [(³√(381√3 – 54√129) – ³√(381√3 + 54√129))i + 2 – ³√(127 + 18√43) – ³√(127 – 18√43)]/6

注意：最後果兩個complex root千祈唔可以reject！因為complex root is also a kind of root！

2007-08-31 20:31:46 補充：
話時話，cheese_cake_leong，究竟你本書嘅果條題目印錯咗啲咩嘢？
不過有一個咁難得嘅機會俾我練三次公式，真係非常多謝你！

1) (x^2+2x+1)/(x^2+5x+4) + (x-1)/(x^2-3x-4) = 0

(x+1)^2/(x+1)(x+4) = - (x-1)/(x+1)(x-4)

(x+1)/(x+4) = - (x-1)/(x+1)(x-4)

(x+1)(x+1)(x-4) = -(x-1)(x+4)

(x^2+2x+1)(x-4) = - (x^2+3x-4)

(x^3+2x^2+x - 4x^2-8x-4) = -x^2-3x+4

x^3-2x^2 -7x-4 = -x^2-3x+4

x^3 -x^2-4x-8=0

2) (x-1)/(x^2-3x-4) = (x-1)/(x+1)(x-4)

Shadow.

(x² + 2x + 1) (x - 1)
------------------ + --------------- = 0
(x² + 5x + 4) (x² - 3x - 4)

(x+1)^2/(x+4)(x+1) + (x-1)/(x-4)(x+1) = 0

(x+1)/(x+4) = - (x-1)/((x-4)(x+1)
(x+1)(x-4)(x+1) = -(x-1)(x+4)
(x^2 +2x +1)(x-4) = -x^2 -3x + 4
x^3 -4x^2 +2x^2 -8x +x -4 = -x^2 -3x +4
x^3 -x^2 -5x -8 = 0

(x-1)
------------
(x²-3x-4)

= (x-1)/(x-4)(x+1)

(

2007-08-24 10:30:24 補充：
Sorry, answer for part 1 should be x^3 -x^2 -4x -8 = 0