Question

Show that $ \left| \begin{matrix}
   b+c & c+a & a+b \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ .

Answer 1

Hint: We first use the row and column operations to simplify the determinant values. We take $ a+b+c $ as the common component from the same row. Then we expand the determinant to find the final value. We use formula of \[\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\].

Complete step-by-step answer:
We need to show that the determinant value will be equal to $ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ .
We can apply row operations on the determinant value without changing the initial form.
So, we take the form of $ {{R}_{1}}^{'}={{R}_{1}}+{{R}_{3}} $ .
We get $ \left| \begin{matrix}
   b+c & c+a & a+b \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|=\left| \begin{matrix}
   a+b+c & b+c+a & a+b+c \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right| $ .
We take common the term $ a+b+c $ from the first row.
 $ \left| \begin{matrix}
   b+c & c+a & a+b \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|=\left( a+b+c \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right| $
Now we simplify further by taking the operations like $ {{C}_{2}}^{'}={{C}_{2}}-{{C}_{1}} $ and $ {{C}_{3}}^{'}={{C}_{3}}-{{C}_{1}} $ .
We get $ \left( a+b+c \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|=\left( a+b+c \right)\left| \begin{matrix}
   1 & 0 & 0 \\
   a+b & c-a & c-b \\
   a & b-a & c-a \\
\end{matrix} \right| $
Now we expand the determinant value through the first row.
So, $ \left| \begin{matrix}
   b+c & c+a & a+b \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|=\left( a+b+c \right)\left[ {{\left( c-a \right)}^{2}}-\left( c-b \right)\left( b-a \right) \right] $ .
We simplify the expression and get
\[\begin{align}
  & \left( a+b+c \right)\left[ {{\left( c-a \right)}^{2}}-\left( c-b \right)\left( b-a \right) \right] \\
 & =\left( a+b+c \right)\left( {{c}^{2}}+{{a}^{2}}-2ac-bc+ac+{{b}^{2}}-ab \right) \\
 & =\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right) \\
\end{align}\]
We also know that \[\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\].
Therefore, $ \left| \begin{matrix}
   b+c & c+a & a+b \\
   a+b & b+c & c+a \\
   a & b & c \\
\end{matrix} \right|={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ .

Note: There are certain operations which we can apply for the problems. We can switch two rows or columns which causes the determinant to switch sign. We can add a multiple of one row to another which causes the determinant to remain the same. We can multiply a row as a constant result in the determinant scaling by that constant.