Ta có : (x+y)2 = 42
=> x2+y2+2xy= 16
=> 2xy= 16-10
=> xy=3
Lại có: x3+y3 = (x+y)3 - 3xy(x+y) = 43-3.3.4=28
xy =\(\frac{x^2+2xy+y^2-\left(x^2+y^2\right)}{2}=\frac{\left(x+y\right)^2-4}{2}=\frac{4^2}{2}-2=6\)
=> x3 + y3 = (x + y)(x2 - xy + y2) = 4(10 - 6) = 4.4 = 16
x+y=4 => (x+y)2=16
x2+2xy+y2=16
10+2xy=16
2xy=6
xy=3
x3+y3=(x+y)(x2+xy+y2)=4*(10-3)=4*7=28
Vậy x3+y3=28
Ta có x2 +y2 =( x+y)2-2xy=1
=> 42-2xy=10
=> -2xy=-6
=> xy=3
Ta lại có : x3+y3=(x+y)3 . 3xy.(x+y)
=43-3.3.4=28
x+y=4 =>(x+y)2 =16
=>x2+2xy+y2=16
=>(x2+y2)+2xy=16
=>10+2xy=16
=>2xy=6
=>xy=3
Ta có hằng đẳng thức:
x3+y3=(x+y).(x2 +xy +y2)
=>x3+y3=4.(10+3)=4.13=42
vậy x3+y3=42
