Question

Factorise the following:
${(a-\frac{1}{2}b+\frac{1}{3}c)}^3-a^3+\frac{1}{8}{b}^3-\frac{1}{27}{c}^3$

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 $a=a$, $b=-\frac{1}{2}b$ and $c=\frac{1}{3}c$, the equation can be factorised as follows:

${(a-\frac{1}{2}b+\frac{1}{3}c)}^3-a^3+\frac{1}{8}{b}^3-\frac{1}{27}{c}^3=3(a-\frac{1}{2}b)(-\frac{1}{2}b+\frac{1}{3}c)(a+\frac{1}{3}c)$

$=3(a-\frac{b}{2})(-\frac{b}{2}+\frac{c}{3})(a+\frac{c}{3})$

Hence, ${(a-\frac{1}{2}b+\frac{1}{3}c)}^3-a^3+\frac{1}{8}{b}^3-\frac{1}{27}{c}^3=3(a-\frac{b}{2})(-\frac{b}{2}+\frac{c}{3})(a+\frac{c}{3})$