<s>[INST] How to prove the determinant? We have to prove the following result without expanding   
$\left|\begin{array}{lll}
      a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
       \end{array}
\right|=(ab+bc+ca)\left|\begin{array}{lll}
      a^2 & a &1 \\
b^2 & b &1\\ c^2 & c &1
       \end{array}
\right|$
Progress :  
$\left|\begin{array}{lll}
      a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
       \end{array}
\right|$
$ 
=\frac{1}{abc}\left|\begin{array}{lll}
      a^3 & a^2 &abc \\
b^3 & b^2 &abc\\ c^3 & c^2 &abc
       \end{array}
\right|$ by $abc\times C_3\rightarrow C_3'$  
$=\left|\begin{array}{lll}
      a^2 & a &bc \\
b^2 & b & ca\\ c^2 & c &ab
       \end{array}
\right|$  by $\frac{1}{a}\times R_1\rightarrow R_1',\frac{1}{b}\times R_2\rightarrow R_2', \frac{1}{c}\times R_3\rightarrow R_3'$ 
$=\left|\begin{array}{ccc}
      a^2+b^2+c^2 & a+b+c &ab+bc+ca\\
b^2 & b & ca\\ c^2 & c &ab
       \end{array}
\right|$ by $ R_1+R_2+R_3\rightarrow R_1' $
Any one tell me how can I show the required result? 
 [/INST] $ \begin{align}\left|\begin{array}{lll}
      a^2 & a &bc+ab+ac+a^2 \\
b^2 & b & ca+ba+b^2+bc\\ c^2 & c &ab+ca+cb+c^2
       \end{array}
\right|\end{align} $ by $(a+b+c) \times C_2 + C_3\rightarrow C_3' $
Then $
 =\left|\begin{array}{lll}
      a^2 & a &bc+ab+ac \\
b^2 & b & ca+ba+bc\\ c^2 & c &ab+ca+cb
       \end{array}
\right| $ by $C_3 - C_1\rightarrow C_3'$
Hence
$\left|\begin{array}{lll}
      a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
       \end{array}
\right|=(ab+bc+ca)\left|\begin{array}{lll}
      a^2 & a &1 \\
b^2 & b &1\\ c^2 & c &1
       \end{array}
\right|$
</s>