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

1 thought on “x3 + y3 – 1 + 3xy, x + y – 1 divide the first term with second​”

1. Answer:

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

   Step-by-step explanation:

   [tex]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)\\\\[/tex]

   DIVIDING BOTH SIDES BY x+y-1

   [tex]\bf \frac{x^3 + y^3 – 1 + 3xy}{x+y-1} =x^2+y^2 -xy+x+y+1[/tex]

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