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Prove the following relation.
\[\left| \begin{matrix}
   {{a}^{2}} & {{a}^{2}}-{{\left( b-c \right)}^{2}} & bc \\
   {{b}^{2}} & {{b}^{2}}-{{\left( c-a \right)}^{2}} & ca \\
   {{c}^{2}} & {{c}^{2}}-{{\left( a-b \right)}^{2}} & ab \\
\end{matrix} \right|=\left( b-c \right)\left( c-a \right)\left( a-b \right)\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\]

Answer
VerifiedVerified
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Hint: Separate the given determinant into two parts by splitting the middle column, that is, write \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] and \[{{\left( b-c \right)}^{2}},{{\left( c-a \right)}^{2}},{{\left( a-b \right)}^{2}}\] in different determinants. Then use operations \[{{C}_{2}}\to {{C}_{2}}+2{{C}_{3}}\]and then proceed.

Here, we have to prove that
\[\left| \begin{matrix}
   {{a}^{2}} & {{a}^{2}}-{{\left( b-c \right)}^{2}} & bc \\
   {{b}^{2}} & {{b}^{2}}-{{\left( c-a \right)}^{2}} & ca \\
   {{c}^{2}} & {{c}^{2}}-{{\left( a-b \right)}^{2}} & ab \\
\end{matrix} \right|=\left( b-c \right)\left( c-a \right)\left( a-b \right)\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\]
Let us consider the given determinant as,
\[D=\left| \begin{matrix}
   {{a}^{2}} & {{a}^{2}}-{{\left( b-c \right)}^{2}} & bc \\
   {{b}^{2}} & {{b}^{2}}-{{\left( c-a \right)}^{2}} & ca \\
   {{c}^{2}} & {{c}^{2}}-{{\left( a-b \right)}^{2}} & ab \\
\end{matrix} \right|\]
Since, we know that
\[\left| \begin{matrix}
   x & a-b & m \\
   y & c-d & n \\
   z & e-f & 0 \\
\end{matrix} \right|=\left| \begin{matrix}
   x & a & m \\
   y & c & n \\
   z & e & 0 \\
\end{matrix} \right|-\left| \begin{matrix}
   x & b & m \\
   y & d & n \\
   z & f & 0 \\
\end{matrix} \right|\]
Therefore, we can apply it to the determinant in the question, D as shown below.
\[D=\left| \begin{matrix}
   {{a}^{2}} & {{a}^{2}} & bc \\
   {{b}^{2}} & {{b}^{2}} & ca \\
   {{c}^{2}} & {{c}^{2}} & ab \\
\end{matrix} \right|-\left| \begin{matrix}
   {{a}^{2}} & {{\left( b-c \right)}^{2}} & bc \\
   {{b}^{2}} & {{\left( c-a \right)}^{2}} & ca \\
   {{c}^{2}} & {{\left( a-b \right)}^{2}} & ab \\
\end{matrix} \right|\]
We know that if any two rows or any two columns of determinant is identical, then the value of determinant is zero.
Therefore, we get
\[D=0-\left| \begin{matrix}
   {{a}^{2}} & {{\left( b-c \right)}^{2}} & bc \\
   {{b}^{2}} & {{\left( c-a \right)}^{2}} & ca \\
   {{c}^{2}} & {{\left( a-b \right)}^{2}} & ab \\
\end{matrix} \right|\]
We know that \[{{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\]
Therefore, we get,
\[D=-\left| \begin{matrix}
   {{a}^{2}} & {{b}^{2}}+{{c}^{2}}-2bc & bc \\
   {{b}^{2}} & {{c}^{2}}+{{a}^{2}}-2ca & ca \\
   {{c}^{2}} & {{a}^{2}}+{{b}^{2}}-2ab & ab \\
\end{matrix} \right|\]
Now, we will use the operation,
\[{{C}_{2}}\to {{C}_{2}}+2{{C}_{3}}\]
We will get,
\[D=-\left| \begin{matrix}
   {{a}^{2}} & {{b}^{2}}+{{c}^{2}}-2bc+2bc & bc \\
   {{b}^{2}} & {{c}^{2}}+{{a}^{2}}-2ca+2ca & ca \\
   {{c}^{2}} & {{a}^{2}}+{{b}^{2}}-2ab+2ab & ab \\
\end{matrix} \right|\]
By cancelling like terms and simplifying, we get,
\[D=-\left| \begin{matrix}
   {{a}^{2}} & {{b}^{2}}+{{c}^{2}} & bc \\
   {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & ca \\
   {{c}^{2}} & {{a}^{2}}+{{b}^{2}} & ab \\
\end{matrix} \right|\]
Now, we will use the operation,
\[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\]
So, we will get
\[D=-\left| \begin{matrix}
   {{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}}+{{c}^{2}} & bc \\
   {{b}^{2}}+{{c}^{2}}+{{a}^{2}} & {{c}^{2}}+{{a}^{2}} & ca \\
   {{c}^{2}}+{{a}^{2}}+{{b}^{2}} & {{a}^{2}}+{{b}^{2}} & ab \\
\end{matrix} \right|\]
By taking \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\] common from \[{{R}_{1}}\], we get
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix}
   1 & {{b}^{2}}+{{c}^{2}} & bc \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   1 & {{a}^{2}}+{{b}^{2}} & ab \\
\end{matrix} \right|\]
Now, we will use the operation,
\[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]
And, \[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\]
We will get,
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix}
   1-1 & \left( {{b}^{2}}+{{c}^{2}} \right)-\left( {{c}^{2}}+{{a}^{2}} \right) & bc-ca \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   1-1 & \left( {{a}^{2}}+{{b}^{2}} \right)-\left( {{c}^{2}}+{{a}^{2}} \right) & ab-ca \\
\end{matrix} \right|\]
\[\Rightarrow D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix}
   0 & {{b}^{2}}-{{a}^{2}} & c\left( b-c \right) \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   0 & {{b}^{2}}-{{c}^{2}} & a\left( b-c \right) \\
\end{matrix} \right|\]
Since, we know that \[\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x-y \right)\left( x+y \right)\], we can apply it to the above determinant and we will get
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix}
   0 & \left( b-a \right)\left( b+a \right) & c\left( b-c \right) \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   0 & \left( b-c \right)\left( b+c \right) & a\left( b-c \right) \\
\end{matrix} \right|\]
By taking \[\left( b-a \right)\] and \[\left( b-c \right)\] common from \[{{R}_{1}}\] and \[{{R}_{3}}\] respectively, we get
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( b-a \right)\left( b-c \right)\left| \begin{matrix}
   0 & b+a & c \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   0 & b+c & a \\
\end{matrix} \right|\]
Let,
\[\Delta =\left| \begin{matrix}
   0 & b+a & c \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   0 & b+c & a \\
\end{matrix} \right|\]
Therefore, we get
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( b-a \right)\left( b-c \right).\Delta ....\left( i \right)\]
We know that determinant value of
\[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-fh \right)-d\left( bi-hc \right)+g\left( bf-ec \right)\]
Therefore determinant value of \[\Delta =\left| \begin{matrix}
   0 & b+a & c \\
   1 & {{c}^{2}}+{{a}^{2}} & ca \\
   0 & b+c & a \\
\end{matrix} \right|\] is
\[0\left[ \left( {{c}^{2}}+{{a}^{2}} \right)\left( a \right)-\left( ca \right)\left( b+c \right) \right]-1\left[ \left( b+a \right)\left( a \right)-\left( c \right)\left( b+c \right) \right]+0\left[ \left( b+a \right)\left( ca \right)-c\left( {{c}^{2}}+{{a}^{2}} \right) \right]\]
Therefore, we get
\[\Delta =0-\left[ \left( ab+{{a}^{2}} \right)-\left( cb+{{c}^{2}} \right) \right]+0\]
\[\Rightarrow \Delta ={{c}^{2}}+bc-ab-{{a}^{2}}\]
We know that
\[{{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)\]
Therefore, we get
\[\Delta =\left( c-a \right)\left( c+a \right)+b\left( c-a \right)\]
By taking \[\left( c-a \right)\] common, we get
\[\Delta =\left( c-a \right)\left( c+a+b \right)\]
By putting the value of \[\Delta \] in equation (i), we get,
\[D=-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( b-a \right)\left( b-c \right)\left( c-a \right)\left( c+a+b \right)\]
Or, \[D=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)\]
Hence Proved.

Note: In these types of questions, always try to first take out the common terms which are the same as terms in RHS, then try to make the terms of each row or column equal to 0 to easily find the determinant value.