Question

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$

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

Answer 1

Let $\Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
Operating $R_1\to R_1-R_3$ and $R_2\to R_2-R_3$, we get

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

Taking common factors (a-c) and (b-c) from $R_1$ and $R_2$ respectively, we get $=(a-c)(b-c)\left|\begin{array}{ccc}0& 1& (a+c)\\ 0& 1& (b+c)\\ 1& c& c^2\end{array}\right|$

Now, expanding corresponding to $C_1$ we get
$=(a-c)(b-c)(b+c-a-c)=(a-b)(b-c)(c-a)=RHS$
Hence proved.

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.