Question

Prove that
$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$
Operating $C_1\to C_1-C_{2}and{C}_2\to C_2-C_3$, we get

$=\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^3-b^3& b^3-c^3& c^3\end{array}\right|=\left|\begin{array}{ccc}0& 0& 1\\ (a-b)& (b-c)& c\\ (a-b)(a^2+ab+b^2)& (b-c)(b^2+bc+c^2)& c^3\end{array}\right|[\because x^3-y^3=(x-y\left)\right(x^2+xy+y^2\left)\right]$

Taking common (a-b) from $C_1$ and $(b-c)$ from $C_2$. We get
$=(a-b)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ a^2+ab+b^2& b^2+bc+c^2& c^3\end{array}\right|$
Now expanding along $R_1$. We get
$=(a-b)(b-c)[1\times (b^2+bc+c^2)-1\times (a^2+ab+b^2\left)\right]=(a-b)(b-c)[b^2+bc+c^2-a^2-ab-b^2]=(a-b)(b-c)[bc-ab+c^2-a^2]=(a-b)(b-c)\left[b\right(c-a{)}_{(}c-a\left)\right(c+a\left)\right]=(a-b)(b-c)(c-a)(a+b+c)=RHS$.
Hence proved.
In this type of questions we only use either row operations or column operations not both at one time.