`7, G=(3x+1)^3-(y-2)^2+(y-1)^3-(x+y)^2`
`\qquadG=27x^3+27x^2+9x+1-y^2+4y-4+y^3-3y^2+3y-1-x^2-2xy-y^2`
`\qquadG=27x^3+y^3+(27x^2-x^2)+(-y^2-3y^2-y^2)+9x+(4y+3y)-2xy+(1-4-1)`
`\qquadG=27x^3+y^3+26x^2-5y^2+9x+7y-2xy-4`
Thay `x=1/3; y=-3` vào G ta có:
`\qquadG=27. (1/3)^3+(-3)^3+26.(1/3)^2-5.(-3)^2+9. 1/3+7.(-3)-2. 1/3. (-3)-4`
`\qquadG=27. 1/27-27+26. 1/9-5.9+3-21+2-4`
`\qquadG=1-27+26/9-45+3-21+2-4=-793/9`
Vậy `G=-793/9` khi `x=1/3; y=-3`