Question

Factorise the following:
${(x-y-z)}^3-x^3+y^3+z^3$

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 $a=x$, $b=-y$ and $c=-z$, the equation $(x-y-z{)}^3-x^3+y^3+z^3$ can be factorised as follows:

$(x-y-z{)}^3-x^3+y^3+z^3=3(x-y\left)\right(-y-z\left)\right(-z+x)=-3(x-y\left)\right(y+z\left)\right(-z+x)=3(x-y\left)\right(y+z\left)\right(z-x)$

Hence, $(x-y-z{)}^3-x^3+y^3+z^3=3(x-y\left)\right(y+z\left)\right(z-x)$