x+y=7 xy=20 3x^3+3y^3=

=3(x^3+y^3)
=3[(x+y)(x^2-xy+y^2)]
=3{(x+y)[(x+y)^2-2xy-xy]}
剩下的不用说了吧

3(x^3+y^3)=3(x+y)(x^2-xy+y^2)=3(x+y)[(x+y)^2-3xy)]
=3*7*(7^2-3*20)=-231

3x^3+3y^3=3(x^3+y^3)=3(x+y)(x^2-xy+y^2)=3(x+y)(x^2+2xy+y^2-3xy)=3(x+y)((x+y)^2-3xy)
把x+y=7，xy=20代入，得原式=-231

运用公式x^3+y^3=(x+y)(x^2-xy+y^2)
所以3x^3+3y^3=3(x^3+y^3)
=3(x+y)(x^2-xy+y^2)
=3(x+y)(x^2+2xy+y^2-3xy)
=3(x+y)((x+y)^2-3xy)
=3*7*（49-60）
=-231