Question

Factorise the following:
${(x+y+z-3)}^3-{(x-1)}^3-{(y-1)}^3-{(z-1)}^3$

Answer 1

Let us consider $x-1+y-1+z-1=x+y+z-3$, therefore, the given problem can be rewritten as:

$\left[\right(x-1)+(y-1)+(z-1){]}^3-(x-1{)}^3-(y-1{)}^3-(z-1{)}^3$

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 $a=(x-1)$, $b=(y-1)$ and $c=(z-1)$, the equation

$\left[\right(x-1)+(y-1)+(z-1){]}^3-(x-1{)}^3-(y-1{)}^3-(z-1{)}^3$ can be factorised as follows:

$\left[\right(x-1)+(y-1)+(z-1){]}^3-(x-1{)}^3-(y-1{)}^3-(z-1{)}^3$

$=3\left(\right(x-1)+(y-1\left)\right)\left(\right(y-1)+(z-1\left)\right)\left(\right(x-1)+(z-1\left)\right)=3(x-1+y-1)(y-1+z-1)(x-1+z-1)$

$=3(x+y-2)(y+z-2)(x+z-2)$

Hence, $(x+y+z-3{)}^3-(x-1{)}^3-(y-1{)}^3-(z-1{)}^3=3(x+y-2)(y+z-2)(z+x-2)$