A F. 2 Maths question - Factorization by using identities

A F. 2 Maths question - Factorization by using identities

m6 + n6

= (m2)3 + (n2)3

= [m2 + n2][(m2)2 - (m2)(n2) + (n2)2] 【Using identity a3+b3=(a+b)(a2-ab+b2)】

= (m2 + n2)(m4 - m2n2 + n4)


2007-01-09 20:45:23 補充：
小小補充：factorizaton (因式分解) 常用到的恆等式有a²-b² = (a-b)(a+b) ..... 最常用(a-b)² = a²-2ab+b²(a+b)² = a²+2ab+b²a³-b³ = (a-b)(a²+ab+b²)a³+b³ = (a+b)(a²-ab+b²)只是代不同的項至 a 和 b。

m^6+n^6
=(m^3)^2+(n^3)^2
=(m^3+in^3)(m^3+in^3)【Using identity a^2+b^2=(a+bi)(a-bi)】i = √-1 (imaginary number)

2007-01-09 20:59:07 補充：
a^2+b^2=a^2-(-b^2)=a^2-(-1*b^2)=(a+b√-1)(a-b√-1)=(a+bi)(a-bi)

2007-01-10 16:10:57 補充：
m^6+n^6=(m^3)^2+(n^3)^2=(m^3+in^3)(m^3-in^3)【Using identity a^2+b^2=(a+bi)(a-bi)】i = √-1 (imaginary number)

A F. 2 Maths question - Factorization by using identities

m6 + n6

= (m2)3 + (n2)3

= [m2 + n2][(m2)2 - (m2)(n2) + (n2)2] 【Using identity a3+b3=(a+b)(a2-ab+b2)】

= (m2 + n2)(m4 - m2n2 + n4)

m^6+n^6+
(m^2)^3 + (n^2)^3
let a = a^2 and b=n^2
(a+b)(a^2-ab+b^2)
=(m^2+n^2)(m^4-m^2 n^2 + n^4)