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  4. The value of the determinant | beginmatrix 0 b3-a3 c3-a3 a3-b3 0 c3-b3 a3-c3 b3-c3 0 endmatrix | is equal to

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Q. The value of the determinant $ \left| \begin{matrix} 0 & {{b}^{3}}-{{a}^{3}} & {{c}^{3}}-{{a}^{3}} \\ {{a}^{3}}-{{b}^{3}} & 0 & {{c}^{3}}-{{b}^{3}} \\ {{a}^{3}}-{{c}^{3}} & {{b}^{3}}-{{c}^{3}} & 0 \\ \end{matrix} \right| $ is equal to

J & K CETJ & K CET 2005

Solution:

$ \left| \begin{matrix} 0 & {{b}^{3}}-{{a}^{3}} & {{c}^{3}}-{{a}^{3}} \\ {{a}^{3}}-{{b}^{3}} & 0 & {{c}^{3}}-{{b}^{3}} \\ {{a}^{3}}-{{c}^{3}} & {{b}^{3}}-{{c}^{3}} & 0 \\ \end{matrix} \right| $
$ =\left| \begin{matrix} 0 & -({{a}^{3}}-{{b}^{3}}) & -({{a}^{3}}-{{c}^{3}}) \\ {{a}^{3}}-{{b}^{3}} & 0 & -({{b}^{3}}-{{c}^{3}}) \\ {{a}^{3}}-{{c}^{3}} & {{b}^{3}}-{{c}^{3}} & 0 \\ \end{matrix} \right| $
$ =({{a}^{3}}-{{b}^{3}})[0+({{b}^{3}}-{{c}^{2}})({{a}^{3}}-{{c}^{3}})] $
$ -({{a}^{3}}-{{c}^{3}})[({{b}^{3}}-{{c}^{3}})({{a}^{3}}-{{b}^{3}})-0] $
$ =({{a}^{3}}-{{b}^{3}})({{b}^{3}}-{{c}^{3}})({{a}^{3}}-{{c}^{3}}) $
$ -({{a}^{3}}-{{c}^{3}})({{b}^{3}}-{{c}^{3}})({{a}^{3}}-{{b}^{3}}) $
$ =0 $