If $Δ_1 =\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix},Δ_2 =\begin{vmatrix}1&bc&a\\1&ca&b\\1&ab&c\end{vmatrix}$, then

$Δ_1+Δ_2=0$

We have,

$Δ_2 =\begin{vmatrix}1&bc&a\\1&ca&b\\1&ab&c\end{vmatrix}$

$⇒Δ_2 =\frac{1}{abc}\begin{vmatrix}a&abc&a^2\\b&abc&b^2\\c&abc&c^2\end{vmatrix}$  Applying $\begin{matrix}R_1 →R_1(a)\\R_2 →R_2(b)\\R_3 →R_3(c)\end{matrix}$

$⇒Δ_2 =\begin{vmatrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{vmatrix}$ [Taking abc common from $C_2$]

$⇒Δ_2 =-\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$  [Interchanging $C_1$ and $C_2$]

$⇒Δ_2 =-\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}=-Δ_1⇒Δ_1+Δ_2=0$.