Question

Using properties of determinants, prove the following:
$\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|=a^3+b^3+c^3-3abc$

Answer 1

Let us take the LHS of the given expression.
$\text{LHS}=\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|$

$\text{Perform}{C}_1\to C_1+C_2+C_3\text{LHS}=\left|\begin{array}{ccc}a+b+c& b& c\\ 0& b-c& c-a\\ 2(a+b+c)& c+a& a+b\end{array}\right|$

$\text{Take}(a+b+c)\text{common from}{C}_1,$
$\text{LHS}=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& b-c& c-a\\ 2& c+a& a+b\end{array}\right|$

$\text{Perform}{R}_3\to R_3-2R_1\text{LHS}=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& b-c& c-a\\ 0& c+a-2b& a+b-2c\end{array}\right|$

Now, expanding the determinant along $C_1,$ we get
$\text{LHS}=(a+b+c)((b-c)(a+b-2c)-(c+a-2b)(c-a))$
$=(a+b+c)(ab+b^2-2bc-ac-bc+2c^2-c^2+ac-ac+a^2+2bc-2ab)$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

We know that
$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$
$\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
$\therefore \text{LHS}=$$a^3+b^3+c^3-3abc$

$\text{Thus, LHS = RHS}$