Question

x3 + y3 - 1 + 3xy, x + y - 1 divide the first term with second​

Answer 1

Answer:

Quotient =x^2+y^2-xy+x+y+1

Step-by-step explanation:

$x^3 + y^3 - 1 + 3xy\\\\=(x+y)^3-3xy(x+y)- 1 + 3xy\\\\=(x+y)^3-1-3xy(x+y)+3xy\\\\=( x+y-1) [ (x+y)^2+(x+y)(1)+1]-3xy( x+y-1)\\\\=(x+y-1) ( x^2+y^2+xy+x+y+1)-3xy( x+y-1)\\\\=(x+y-1) ( x^2+y^2+2xy+x+y+1-3xy)\\\\ Or\\\\x^3 + y^3 - 1 + 3xy=(x+y-1) ( x^2+y^2-xy+x+y+1)$

DIVIDING BOTH SIDES BY x+y-1

$\bf \frac{x^3 + y^3 - 1 + 3xy}{x+y-1} =x^2+y^2 -xy+x+y+1$