邊個識計呢條數?

x6 + 2x4y2 + y4x2 - 4

= x2 (x4 + 2x2y2 + y4) - 4

= x2 [(x2)2 + 2(x2)(y2) + (y2)2] - 4

= x2[(x2) + (y2)]2 - 4 【根據恆等式 (a+b)2 = a2 + 2ab + b2】

= x2(x2 + y2)2 - 4

= [x(x2 + y2)]2 - 4

= (x3 + xy2)2 - 4

= (x3 + xy2)2 - 22

= [(x3 + xy2) - 2][(x3 + xy2) + 2] 【根據恆等式 a2-b2 = (a-b)(a+b)】

= (x3 + xy2 - 2)(x3 + xy2 + 2)

2006-12-23 11:46:01 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+b²

x^6+2x^4y^2+y^4x^2-4
=X^2 (x^4 + 2X^2 y^2 +y^4) -4
=x^2 (x^2 + y^2)^2 -4
= [ x(x^2 + y^2)] ^2 -4
= [ x( x^2 + y^2] ^2 - 2^2
= [ x(x^2 + y^2) +2] [ x(x^2 +y^2) -2]
= (x^3 + xy^2 +2) ( x^3 +xy^2 -2)

(as your answer shown)

x^6+2x^4y^2+y^4x^2 –4
= x^2(x^4+2x^2y^2+y^4)-4 by completing square (a^2+b^2)^2= a^2+2ab+b^2
= x^2[(x^2+y^2)^2] –4
= [x (x^2+y^2)^2]^2 –2^2 by different of 2 squares
= [x (x^2+y^2)^2+2][ x (x^2+y^2)^2-2] simplify
= (x^3+xy^2+2)(x^3+xy^2-2)