Question

Factorize $a^3+b^3+c^3-3abc$.

Answer 1

Step 1: Factorize the given expression

$a^3+b^3+c^3-3abc$

use the identity

$$\begin{gathered} {(x+y)}^3=x^3+y^3+3xy(x+y) \\ \Rightarrow {(x+y)}^3-3xy(x+y)=x^3+y^3 \end{gathered}$$

put $x=b,y=c$

$\therefore$$a^3+\left[b^3+c^3\right]-3abc$

$$\begin{gathered} =a^3+\left[{(c+b)}^3-3cb(c+b)\right]-3abc \\ =a^3+{(b+c)}^3-3bc(a+b+c) \end{gathered}$$

Step 2: Factorize further

using the identity $x^3+y^3=(x+y)(x^2+y^2-xy)$

$x=a,y=b+c$

$a^3+{(b+c)}^3=(a+b+c)\left[a^2+{(b+c)}^2-a(b+c)\right]$

$$\begin{gathered} \therefore a^3+{(b+c)}^3-3bc(a+b+c)=(a+b+c)\left[a^2+{(b+c)}^2-a(b+c)\right]-3bc(a+b+c) \\ =(a+b+c)\left[a^2+b^2+c^2+2bc-ab-ac-3bc\right] \\ \Rightarrow a^3+{(b+c)}^3-3bc(a+b+c)=(a+b+c)\left[a^2+b^2+c^2-bc-ab-ac\right] \end{gathered}$$

Hence the factorized form of the polynomial $a^3+b^3+c^3-3abc$ is $(a+b+c)$$(a^2+b^2+c^2-bc-ab-ac)$.