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Mathematics Question on Determinants

Without expanding the determinant, prove that
∣aa2bc bb2ca cc2ab∣=∣1a2a3 1b2b3 1c2c3∣\begin{vmatrix} a &a^2 &bc \\\ b&b^2 &ca \\\ c&c^2 &ab \end{vmatrix}=\begin{vmatrix} 1 &a^2 &a^3 \\\ 1&b^2 &b^3 \\\ 1&c^2 &c^3 \end{vmatrix}

Answer

L.H.S= ∣aa2bc bb2ca cc2ab ∣\begin{vmatrix} a& a^2& b_c\\\ b & b^2 & ca \\\ c& c^2& ab \\\ \end{vmatrix}
=1abc∣a2a3abc b2b3abc c2c3abc ∣[R1→aR1,R2→bR2,R3→cR3]=\frac{1}{abc}\begin{vmatrix} a^2& a^3& abc\\\ b^2 & b^3 & abc \\\ c^2& c^3& abc \\\ \end{vmatrix} [R_1\rightarrow{aR_1,R_2}\rightarrow{bR_2,R_3}\rightarrow{cR_3}]
=abcabc∣a2a31 b2b31 c2c31 ∣=∣a2a31 b2b31 c2c31 ∣=\frac{abc}{abc}\begin{vmatrix} a^2& a^3& 1\\\ b^2 & b^3 & 1 \\\ c^2& c^3& 1 \\\ \end{vmatrix} =\begin{vmatrix} a^2& a^3& 1\\\ b^2 & b^3 & 1 \\\ c^2& c^3& 1 \\\ \end{vmatrix} [Taking out factor ABC from C,]
= ∣1a2a3 1b2b3 1c2c3 ∣\begin{vmatrix} 1& a^2& a^3\\\ 1 & b^2& b^3 \\\ 1& c^2& c^3 \\\ \end{vmatrix} [Applying C1→C1C_1\rightarrow{C_1} C2→C_2\rightarrow
=R.H.S
Hence, the given result is proved.