Question

Mark the correct alternative in each of the following: The factors of $x^3-1+y^3+3xy$, are

• A. $(x-1+y)$
  $(x^2+1+y^2+x+y-xy)$
• B. $(x-1+y)$
  $(x^2-1-y^2+x+y+xy)$
• C. $3(x+y-1)(x^2+y^2-1)$
• D. $(x+y+1)$
  $(x^2+y^2+1-xy-x-y)$

Answer 1

The correct option is A $(x-1+y)$

$(x^2+1+y^2+x+y-xy)$
Given: $x^3-1+y^3+3xy$

$\therefore x^3-1+y^3+3xy$

$=(x{)}^3+(-1{)}^3+(y{)}^3-3\times x\times (-1)\times y$

$=(x-1+y)\left[\right(x{)}^2+(-1{)}^2+(y{)}^2-x\times (-1)-(-1)\times \left(y\right)-x\times y]$

$[\because a^3+b^3+c^2-3abc=(a+b+c\left)\right(a^2+b^2+c^2-ab-bc-ca\left)\right]$

$=(x-1+y)(x^2+1+y^2+x+y-xy)$

Hence, option $\left(A\right)$ is the correct answer.