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  • Using properties of determinants, prove the following: <img alt="\begin{vmatrix} a & b & c\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix} = a^3+b^3+c^3-3abc" src="https://entrancecorner.codecogs.com/gif.latex?%5C

Using properties of determinants, prove the following:

\begin{vmatrix} a & b & c\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix} = a^3+b^3+c^3-3abc

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} a & b & c\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix} = a^3+b^3+c^3-3abc

LHS = \begin{vmatrix} a & b & c\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix}

Appyling R_1\rightarrow R_1 + R_3, we get

                \begin{vmatrix} a+b+c & a+b+c & a+ b +c\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix}

Taking (a+b+c) from R1.

               =(a+b+c) \begin{vmatrix} 1 & 1 & 1\\ a-b & b-c & c- a\\ b+c & c+a & a + b \end{vmatrix}

Appyling C_2\rightarrow C_2 - C_1 \& C_3\rightarrow C_3 - C_1

            =(a+b+c) \begin{vmatrix} 1 & 0 & 0\\ a-b & 2b-c-a & c- 2a+ b\\ b+c & a-b & a-c \end{vmatrix}

Now Expanding along R1,  we get

            \\= (a + b+ c)[(1)\{(2b-c-a)(a-c)-(a-b)(c-2a-b)\}]\\ =(a+b+c)[(2ba -2bc -ac + c^2 -a^2 + ac)- (ac-2a^2+ab-bc+2ab -b^2)] \\= (a+b+c)[2ba -2bc +c^2 -a^2-ac +2a^2 -ab +bc -2ab +b^2] \\= (a+b+c)[a^2 +b^2 +c^2 -ab-bc-ca] \\ = a^3 +b^3 +c^3 - 3abc = RHS

Hence proved.

Posted by

Ravindra Pindel

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