Tìm đa thức V sao cho

V + 4y³ – 2xy² + x²y – 9 = 4y³ – 3.

P + Q = 4x²y² – 3xy³ + 5x³y – xy + 2x – 3 – 4x²y² – 4xy³ – x³y + xy + y + 1

= (4x²y² – 4x²y²) + (– 3xy³ – 4xy³) + (5x³y – x³y) + (– xy + xy) + 2x + y + (–3 + 1)

= ‒7xy³ + 4x³y + 2x + y ‒ 2.

P ‒ Q = 4x²y² – 3xy³ + 5x³y – xy + 2x – 3 ‒ (–4x²y² – 4xy³ – x³y + xy + y + 1)

= 4x²y² – 3xy³ + 5x³y – xy + 2x – 3 + 4x²y² + 4xy³ + x³y ‒ xy ‒ y ‒ 1

= (4x²y² + 4x²y²) + (– 3xy³ + 4xy³) + (5x³y + x³y) + (– xy ‒ xy) + 2x ‒ y + (–3 ‒ 1)

= 8x²y² + xy³ + 6x³y ‒ 2xy + 2x ‒ y ‒ 4.

Ta có:

U – 3x²y + 2xy² – 5y³ = 2xy² – xy + 1

Nên U = 2xy² – xy + 1 + 3x²y ‒ 2xy² + 5y³

           = (2xy² ‒ 2xy²) – xy + 3x²y + 5y³ + 1

           = ‒xy + 3x²y + 5y³ + 1.