Question

Factorise the following:
${(a-b-1)}^3-a^3+b^3+1$

Answer 1

We know the identity: $(a+b+c{)}^3-a^3-b^3-c^3=3(a+b\left)\right(b+c\left)\right(c+a)$

Using the above identity taking $a=a$, $b=-b$ and $c=-1$, the equation $(a-b-1{)}^3-a^3+b^3+1$ can be factorised as follows:

$(a-b-1{)}^3-a^3+b^3+1=3(a-b\left)\right(-b-1\left)\right(-1+a)=-3(a-b\left)\right(b+1\left)\right(-1+a)=3(a-b\left)\right(b+1\left)\right(1-a)$

Hence, $(a-b-1{)}^3-a^3+b^3+1=-3(a-b\left)\right(b+1\left)\right(a-1)$