唔識做功課6

(x + y)⁶ - (x - y)⁶
= ( (x + y)² )³ - ( (x - y)² )³
= ( (x + y)² - (x - y)² ) ( (x + y)⁴+ (x + y)² (x - y)² + (x - y)⁴)
= 4xy ( (x + y)⁴+ (x + y)² (x - y)² + (x - y)⁴)
= 4xy ( (x + y)⁴+ 2(x + y)² (x - y)² + (x - y)⁴ - (x + y)² (x - y)² )
= 4xy ( ( (x + y)² + (x - y)² )² - (x + y)² (x - y)² )
= 4xy ( (2x² + 2y²)² - (x + y)² (x - y)² )
= 4xy ( (2x² + 2y²)² - (x² - y²)² )
= 4xy ( 2x² + 2y² - (x² - y²) ) ( 2x² + 2y² + x² - y² )
= 4xy ( x² + 3y² ) ( 3x² + y² )


2012-10-19 15:08:19 補充：
Alternatively :

(x + y)⁶ - (x - y)⁶

= ( (x + y)³ )² - ( (x - y)³ )²

= ( (x + y)³ - (x - y)³ ) ( (x + y)³ + (x - y)³ )

= (x + y - (x - y)) ( (x + y)² + (x + y)(x - y) + (x - y)² ) (x + y + x - y)( (x + y)² - (x + y)(x - y) + (x - y)² )

2012-10-19 15:08:25 補充：
= 2y (x² + 2xy + y² + x² - y² + x² - 2xy + y²) 2x (x² + 2xy + y² - (x² - y²) + x² - 2xy + y²)

= 2y (3x² + y²) 2x (x² + 3y²)

= 4xy ( x² + 3y² ) ( 3x² + y² )

2012-10-19 15:12:27 補充：
(x + y)⁴+ (x + y)² (x - y)² + (x - y)⁴ 加 (x + y)² (x - y)²
變 (x + y)⁴+ 2(x + y)² (x - y)² + (x - y)⁴
可拆成平方 ( (x + y)² + (x - y)² )²

因為加了 (x + y)² (x - y)² 所以要減番 (x + y)² (x - y)² ,

於是係 ( (x + y)² + (x - y)² )² - (x + y)² (x - y)²
又可用平方差分解了~

(x + y)⁶ - (x - y)⁶

= ( (x + y)² )³ - ( (x - y)² )³

= ( (x + y)² - (x - y)² ) ( (x + y)⁴+ (x + y)² (x - y)² + (x - y)⁴)

= 4xy ( (x + y)⁴+ (x + y)² (x - y)² + (x - y)⁴)

= 4xy ( (x + y)⁴+ 2(x + y)² (x - y)² + (x - y)⁴ - (x + y)² (x - y)² )

= 4xy ( ( (x + y)² + (x - y)² )² - (x + y)² (x - y)² )

= 4xy ( (2x² + 2y²)² - (x + y)² (x - y)² )

= 4xy ( (2x² + 2y²)² - (x² - y²)² )

= 4xy ( 2x² + 2y² - (x² - y²) ) ( 2x² + 2y² + x² - y² )

= 4xy ( x² + 3y² ) ( 3x² + y² )





先將6次變成a^3-b^3的公式 a^3-b^3 =(a+b)(a^2-ab+b^2)

((a)^2)^3=a^2*3

之後再抽共有因子

這D題有好多種方法計

唔知你有冇學過呢種既因式解既方法,利用nCr
(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6
(x-y)^6=x^6 - 6x^5y + 15x^4y^2 - 20x^3y^3 + 15x^2y^4 - 6xy^5 + y^6
(x+y)^6 - (x-y)^6 =12x^5y+40x^3y^3 +12xy^5

我唔明第四個等於係如何得出來,可否再解釋,感激！感激！