desenvolvendo em x os polinomios

(x^3+ x^2.3y+ x.3y^2+ y^3-1)

**********************************=

(x^2 + x(2y+1)+ y^2+y+1)

#############################

colocando só os coeficientes de x

1,3y,3y^2,y^3-1 div por (1,2y+1,y^2+y+1)

empregando Briot-Rufini na divisão de polinomios

o divisor (1,2y+1,y^2+y+1)~~~>(2y+1,y^2+y+1)*(-1)~~>

(-2y-1,-y^2-y-1)

****************************************

1,...3y,......3y^2,......y^3-1

....-2y-1,...-y^2-y-1,

**********************

.....y-1,.....2y^2-y-1,...y^3-1

..............-2y^2+y+1,-y^3+1

****************************

..................0,...........,,...0= Resto

*****quociente 1,y-1,ou x+ y-1..*****

************************************

operações feitas

1*(-2y-1,-y^2-y-1)= -2y-1,-y^2-y-1

(y-1)*(-2y-1,-y^2-y-1)= -2y^2+y+1,-y^3+1

***********************************************

verificação ligeira

(1+3y+3y^2+y^3-1) = (1+2y+1+y^2+y+1)*(1+y-1)+

(0 +0)

=(3y+3y^2+y^3)= (3+3y+y^2)*(y) +0

(x + y)³ = (x + y)².(x + y)

(x + y)³ = (x² + 2xy + y²).(x + y)

(x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³

(x + y)³ = x³ + 3x²y + 3xy² + y³

(x + y)³ = x³ + y³ + 3x²y + 3xy²

x³ + y³ = (x + y)³ - 3x²y - 3xy² ← memorize this result as (1)

(a - b)³ = (a - b)²(a - b)

(a - b)³ = (a² - 2ab + b²)(a - b)

(a - b)³ = a³ - a²b - 2a²b + 2ab² + ab² - b³

(a - b)³ = a³ - b³ - 3a²b + 3ab²

(a - b)³ = (a³ - b³) - 3ab(a - b)

a³ - b³ = (a - b)³ + 3ab(a - b)

a³ - b³ = (a - b)².(a - b) + 3ab(a - b)

a³ - b³ = (a - b).[(a - b)² + 3ab]

a³ - b³ = (a - b).[a² - 2ab + b² + 3ab]

a³ - b³ = (a - b).(a² + ab + b²) → suppose that: b = 1

a³ - 1 = (a - 1).(a² + a + 1) ← memorize this result as (2)

Your expression

= [x³ + y³ + 3x²y + 3xy² - 1] / [x² + 2xy + y² + x + y + 1]

= [(x³ + y³) + 3x²y + 3xy² - 1] / [x² + 2xy + y² + x + y + 1] → recall (1)

= [(x + y)³ - 3x²y - 3xy² + 3x²y + 3xy² - 1] / [x² + 2xy + y² + x + y + 1]

= [(x + y)³ - 1] / [x² + 2xy + y² + x + y + 1]

= [(x + y)³ - 1] / [(x² + 2xy + y²) + x + y + 1]

= [(x + y)³ - 1] / [(x + y)² + x + y + 1]

= [(x + y)³ - 1] / [(x + y)² + (x + y) + 1] → let: a = (x + y)

= [a³ - 1] / [a² + a + 1] → recall (2)

= [(a - 1).(a² + a + 1)] / [a² + a + 1] → you can simplify by (a² + a + 1)

= a - 1 → recall: a = (x + y)

= x + y - 1

((x + y)^3 - 1) / ((x + y)^2 + (x + y) + 1)

u = x + y

(u^3 - 1) / (u^2 + u + 1)

Caso de factoreo de diferencia de cubos

u^3 - v^3 = (u - v) (u^2 + u v + v^2)

Con v = 1

u^3 - 1 = (u - 1) (u^2 + u + 1)

Deducimos

(u^3 - 1)/(u^2 + u + 1) = u - 1

en nuestro caso

((x + y)^3 - 1) / ((x + y)^2 + (x + y) + 1) = (x + y) - 1

***********************************************************