Ta có: x + y = 5
=> (x + y )2 = 25
=> x2 + 2xy + y2 = 25
=> 13 + 2xy = 25 ( vì x2 + y2 = 13)
=> 2xy = 12
=> xy = 6
Ta lại có: x3 + y3 = (x + y)( x2 - xy + y2) = 5.(13 - 6) = 35
Vậy ......................
\(( x- y)^2 = 5^2\)
\(=> x^2 - 2xy + y^2 = 25 \)
\(=> 15 - 2xy = 25 \)
\(=> 2xy = -10 \)
\(=> xy = -5 \)
\(x^3 - y^3 = ( x- y)(x^2+xy+y^2) = 5.(15 - 5 ) = 5.10 = 50\)
Ta có: \(x+y=5\)
\(\Rightarrow\left(x+y\right)^2=25\)
\(\Rightarrow x^2+2xy+y^2=25\)
\(\Rightarrow2xy=12\)
\(\Rightarrow xy=6\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=5.\left(13-6\right)\)
\(=5.7=35\)
Vậy \(x^3+y^3=35\)
