Question

Factorise $a^3-b^3+1+3ab$

• A. $(a-b+1)($ $a^2+b^2+ab-a+b+1)$
• B. $(a-b+1)($ $a^2+b^2+ab+a+b+1)$
  
• C. $(a-b-1)($ $a^2+b^2+ab-a+b+1)$
  
• D. $(a-b+2)($ $a^2+b^2+ab-a+b+1)$
  

Answer 1

The correct option is A $(a-b+1)($ $a^2+b^2+ab-a+b+1)$

We factorize the given expression $a^3-b^3+1+3ab$ as shown below:

$a^3-b^3+1+3ab=a^3+(-b{)}^3+1^3+3(a\times -b\times 1)=(a+(-b)+1\left)\right(a^2+(-b{)}^2+1^2-a(-b)-(-b\left)\right(1)-1(a\left)\right)(\because x^3+y^3+z^3+3abc=(x+y+z\left)\right(x^2+y^2+z^2-xy-yz-zx\left)\right)=(a-b+1)(a^2+b^2+1+ab+b-a)$

Hence, $a^3-b^3+1+3ab=(a-b+1)(a^2+b^2+ab-a+b+1)$