Ta có:
x+y=6
=> (x+y)2 = 36
=> x2 +2xy+ y2 = 36
=>20+2xy =36
=> 2xy = 16
=> xy =8
Ta lại có:
x3+y3= (x+y). ( x2 + xy +y2)
= 6 . (20 + 8)
= 120 + 48
= 168
Vậy x3+y3=168
Ta có:
x+y=6
=> (x+y)2 = 36
=> x2 +2xy+ y2 = 36
=>20+2xy =36
=> 2xy = 16
=> xy =8
Ta lại có:
x3+y3= (x+y). ( x2 + xy +y2)
= 6 . (20 + 8)
= 120 + 48
= 168
Vậy x3+y3=168