F6 Maths 74

(1/x³+1/y³)/(1/x+1/y)=[(1/x)³+(1/y)³]/(1/x+1/y)={(1/x+1/y)[(1/x)²-(1/x)(1/y)+(1/y)²]}/(1/x+1/y) ***=1/x²-1/(xy)+1/y²
***Remembera³+b³=(a+b)(a²-ab+b²).

method1:
put u=1/x,v=1/y
so
(1/x^3+1/y^3)/(1/x+1/y)
=(u^3+v^3)/(u+v)
=(u+v)(u^2-uv+v^2)/(u+v)
=u^2-uv+v^2
=1/x^2-1/xy+1/u^2

method2:
(1/x^3+1/y^3)/(1/x+1/y)
=(1/x+1/v)(1/x^2-1/xy+1/u^2)/(1/x+1/y)
=1/x^2-1/xy+1/u^2


2011-11-27 21:52:47 補充：
extra information:
x^3+y^3=(x+y)(x^2-xy+y^2)
x^3-y^3=(x-y)(x^2+xy+y^2)

the first formula is used by this question!! plz remember the above formulas

(1/x^3 + 1/y^3) / (1/x + 1/y)
= [1/(xy)^3] [x^3 + y^3] / [1/xy] [x + y]
= [1/(xy)^2] [(x+y)(x^2 - xy + y^2)] / [x + y]
= (x^2 - xy + y^2) / (xy)^2