\(x^3+y^3=108\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=108\)
\(\Rightarrow x^2-xy+y^2=18\left(1\right)\)
\(x+y=6\Rightarrow\left(x+y\right)^2=36\Leftrightarrow x^2+2xy+y^2=36\left(2\right)\)
Lấy (2)-(1) ta được 3xy=18=>xy=6
Ta có:
x3+y3= ( x+y)(x2- xy+y2) =108
\(\Leftrightarrow\) 6.(x2 -xy + y2) = 108
\(\Leftrightarrow\) x2-xy + y2 = \(\frac{108}{6}\) =18