求解4，5大題，需要過程?

(4)
x³ - 2x - 4 = ( x - α )( x - β )( x - γ ) = 0
比較常數項得 :
- 4 = - αβγ
αβγ = 4 ..... (1式)

比較二次項得 :
0 = - ( α + β + γ )
α + β + γ = 0 ..... (2式)

①
( α³ - 4 )( β³ - 4 )( γ³ - 4 )
= 2α * 2β * 2γ , 因為 x³ - 4 = 2x
= 8 * αβγ
= 8 * 4 , 由(1式)
= 32 ..... Ans

②
α( 1/β + 1/γ ) + β( 1/γ + 1/α ) + γ( 1/α + 1/β )
= α( β + γ )/( βγ ) + β( α + γ )/( αγ ) + γ( α + β )/( αβ )
= α( - α )/( βγ ) + β( - β )/( αγ ) + γ( - γ )/( αβ ) , 由 (2式)
= - [ α²/( βγ ) + β²/( αγ ) + γ²/( αβ ) ]
= - ( α³ + β³ + γ³ )/( αβγ )
= - (1/4)( α³ + β³ + γ³ ) , 由(1式)
= - (1/4)( 2α+4 + 2β+4 + 2γ+4 ) , 因為 x³ = 2x + 4
= - (1/2)( α + β + γ ) - 3
= 0 - 3 , 由 (2式)
= - 3 ..... Ans

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(5)
①
以下反覆利用綜合除法, 將多項式的展開點由 x = 0 轉成 x = 1 :
2x³ + x² - x - 7
= (x-1)(2x²+3x+2) - 5
= (x-1)[ (x-1)(2x+5) + 7 ] - 5
= (x-1){ (x-1)[ 2(x-1) + 7 ] + 7 } - 5
= (x-1)[ 2(x-1)² + 7(x-1) + 7 ] - 5
= 2(x-1)³ + 7(x-1)² + 7(x-1) - 5

即 2x³ + x² - x - 7 = 2(x-1)³ + 7(x-1)² + 7(x-1) - 5

令 u = x - 1 , a = α - 1 , b = β - 1 , c = γ - 1
因為 α 是 2x³ + x² - x - 7 = 0 其中一根,
所以 0 = 2α³ + α² - α - 7 = 2(α-1)³ + 7(α-1)² + 7(α-1) - 5 = 2a³ + 7a² + 7a - 5
因此 a 是 2u³ + 7u² + 7u - 5 = 0 其中一根.
同理, b, c 也是 2u³ + 7u² + 7u - 5 = 0 的根.
因此, 2u³ + 7u² + 7u - 5 = 2( u - a )( u - b )( u - c ) = 0

比較常數項得 :
- 5 = - 2abc
abc = 5/2

比較一次項得 :
7 = 2( ab + ac + bc )
ab + ac + bc = 7/2

1/(α-1) + 1/(β-1) + 1/(γ-1)
= 1/a + 1/b + 1/c
= ( bc + ac + ab ) / ( abc )
= (7/2) / (5/2)
= 7/5 ..... Ans

②
將展開點由 x = 0 轉成 x = - 1 :
2x³ + x² - x - 7
= (x+1)(2x²-x) - 7
= (x+1)[ (x+1)(2x-3) + 3 ] - 7
= (x+1){ (x+1)[ 2(x+1) - 5 ] + 3 } - 7
= (x+1)[ 2(x+1)² - 5(x+1) + 3 ] - 7
= 2(x+1)³ - 5(x+1)² + 3(x+1) - 7

令 u = x + 1 , a = α + 1 , b = β + 1 , c = γ + 1
則 a, b, c 是 2u³ - 5u² + 3u - 7 = 0 的根. ( 詳細推導方式請仿照① )
因此, 2u³ - 5u² + 3u - 7 = 2( u - a )( u - b )( u - c ) = 0
比較各項係數可得 :
a + b + c = 2.5
ab + ac +bc = 1.5
abc = 3.5

以 α/(α+1) , β/(β+1) , γ/(γ+1) 為根的三次方程式為 :
0
= [ x - α/(α+1) ] * [ x - β/(β+1) ] * [ x - γ/(γ+1) ]
= [ x - (a-1)/a ] * [ x - (b-1)/b ] * [ x - (c-1)/c ] , 因為 a = α + 1 , b = β + 1 , c = γ + 1
= ( x - 1 + 1/a )( x - 1 + 1/b )( x - 1 + 1/c )
= ( y + 1/a )( y + 1/b )( y + 1/c ) , 令 y = x - 1
= (1/a)( ay + 1 ) * (1/b)( by + 1 ) * (1/c)( cy + 1 )
= [ 1/(abc) ] * ( ay + 1 )( by + 1 )( cy + 1 )
= ( 1 / 3.5 ) * [ abcy³ + (ab+ac+bc)y² + (a+b+c)y + 1 ]
= ( 2 / 7 ) * ( 3.5y³ + 1.5y² + 2.5y + 1 )
= ( 1 / 7 ) * ( 7y³ + 3y² + 5y + 2 )
= ( 1 / 7 ) * [ 7(x-1)³ + 3(x-1)² + 5(x-1) + 2 ] , 代回 y = x - 1
= ( 1 / 7 ) * [ 7(x³-3x²+3x-1) + 3(x²-2x+1) + 5x-5 + 2 ]
= ( 1 / 7 ) * ( 7x³ - 18x² + 20x - 7 )

故此三次方程式為 :
( 1 / 7 ) * ( 7x³ - 18x² + 20x - 7 ) = 0
即 7x³ - 18x² + 20x - 7 = 0 ..... Ans