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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?

amWhy
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Jacob
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1 Answers1

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$ \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 $\left|\begin{array}{lll} a^3 & a^2 &1 \\ b^3 & b^2 &1\\ c^3 & c^2 &1 \end{array} \right|=(ab+bc+ca)\left|\begin{array}{lll} a^2 & a &1 \\ b^2 & b &1\\ c^2 & c &1 \end{array} \right|$

sn24
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