Cách 2 :
\(x+y=5\\ \Rightarrow\left(x+y\right)^2=25\\ \Rightarrow x^2+y^2+2xy=25\\ \Rightarrow2xy=2\\ \Rightarrow xy=1\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.1.5\\ =125-15=110\)
Ta có:
x2+y2=27
(x+y)2-2xy=27
25-2xy=27
xy=-1
Ta có:
x+y=5
x2+y2=27
=>(x+y)(x2+y2)=135
x3+y3+x2y+xy2=135
x3+y3+xy(x+y)=135
x3+y3-5=135
x3+y3=140
ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=25-2xy=27\Leftrightarrow xy=-1\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=5\left(27+1\right)=140\)
vậy \(x^3+y^3=140\)
