Question

Using properties of determinants, prove the following:

${\displaystyle \left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|=a^3+b^3+c^3-3abc}$.

Answer 1

Δ = ${\displaystyle \left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|}$

 

= ${\displaystyle \left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ a+b+c& c+a+b& a+b+c\end{array}\right| ...[\text{Applying}{R}_3\to R_3+R_1]}$ 

 

= ${\displaystyle (a+b+c) \left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ 1& 1& 1\end{array}\right| ...\left[\text{Taking}(a+b+c)\text{common}\right]}$

 

= ${\displaystyle (a+b+c) \left|\begin{array}{ccc}a& b& c\\ -b& -c& -a\\ 1& 1& 1\end{array}\right| ...[\text{Applying}{R}_2\to R_2-R_1]}$

 

= ${\displaystyle (a+b+c)\left|\begin{array}{ccc}a-c& b-c& c\\ -b+a& -c+a& -a\\ 0& 0& 1\end{array}\right| ...[C_1\to C_1-C_3 \text{and} C_2\to C_2-C_3]}$

 

= ${\displaystyle (a+b+c)\left[\{(a-c)(a-c)-(b-c)(a-b)\}\right]}$

 

= ${\displaystyle (a+b+c) ...[\{{(a-c)}^2-(ab-ac-b^2+bc)]}$

 

= ${\displaystyle (a+b+c) [(a-c^2-(ab-ac-b^2+bc)]}$

 

= ${\displaystyle (a+b+c) (a^2+b^2+c^2-ab-bc-ca)}$

 

= ${\displaystyle a^3+b^3+c^3-3abc}$