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Prove the following
\[\left| {\begin{array}{*{20}{c}}
  {{{\left( {b + c} \right)}^2}}&{{a^2}}&{{a^2}} \\
  {{b^2}}&{{{(c + a)}^2}}&{{b^2}} \\
  {{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right| = 2abc{(a + b + c)^3}\]

Answer Verified Verified
Hint: Here we perform various operations on rows and columns of determinant to make it simple.

Taking the left-hand side of the questions and solving it further so that the left-hand side will become
equal to the right-hand side. So,
$ \Rightarrow $L. H. S =\[\left| {\begin{array}{*{20}{c}}
  {{{\left( {b + c} \right)}^2}}&{{a^2}}&{{a^2}} \\
  {{b^2}}&{{{(c + a)}^2}}&{{b^2}} \\
  {{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|\]
Now, to simplify the determinant we will do various operations so that the determinant becomes
easy and we can expand the determinant without any error. Expanding the determinant can be done
before simplification but, it will make the solution tedious and complicated. So, we will simplify the
determinant and then expand it. So, for simplification we will first subtract the column \[{C_3}\]from
the column ${C_1}$.
Applying ${C_1} \to {C_1} - {C_3}$
$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}
  {{{(b + c)}^2} - {a^2}}&{{a^2}}&{{a^2}} \\
  {{b^2} - {b^2}}&{{{(c + a)}^2}}&{{b^2}} \\
  {{c^2} - {{(a + b)}^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$
Also applying ${C_2} \to {C_2} - {C_3}$
$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}
  {{{(b + c)}^2} - {a^2}}&{{a^2} - {a^2}}&{{a^2}} \\
  {{b^2} - {b^2}}&{{{(c + a)}^2} - {b^2}}&{{b^2}} \\
  {{c^2} - {{(a + b)}^2}}&{{c^2} - {{(a + b)}^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ ……………... (1)
Taking \[(a + b + c)\] common from column ${C_1}$and column \[{C_2}\]
$ \Rightarrow $L. H. S = ${(a + b + c)^2}\left| {\begin{array}{*{20}{c}}
  {b + c - a}&0&{{a^2}} \\
  0&{c + a - b}&{{b^2}} \\
  {c - (a + b)}&{c - (a + b)}&{{{(a + b)}^2}}
\end{array}} \right|$
Also, applying ${R_3} \to {R_3} - {R_1} - {R_2}$
$ \Rightarrow $L. H. S = \[{(a + b + c)^2}\left| {\begin{array}{*{20}{c}}
  {b + c - a}&0&{{a^2}} \\
  0&{c + a - b}&{{b^2}} \\
  { - 2b}&{ - 2a}&{2ab}
\end{array}} \right|\]
Now, multiply and divide column ${C_1}$ by $a$ and column \[{C_2}\] by $b$.
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}
  {ab + ac - {a^2}}&0&{{a^2}} \\
  0&{bc + ab - {b^2}}&{{b^2}} \\
  { - 2ab}&{ - 2ab}&{2ab}
\end{array}} \right|\]
Now, doing ${C_1} \to {C_1} + {C_3}$and ${C_2} \to {C_2} + {C_3}$
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}
  {ab + ac - {a^2} + {a^2}}&{{a^2}}&{{a^2}} \\
  {{b^2}}&{bc + ab - {b^2} + {b^2}}&{{b^2}} \\
  { - 2ab + 2ab}&{ - 2ab + 2ab}&{2ab}
\end{array}} \right|\]
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}
  {ab + ac}&{{a^2}}&{{a^2}} \\
  {{b^2}}&{bc + ab}&{{b^2}} \\
  0&0&{2ab}
\end{array}} \right|\]
Now, our determinant has become simple. So, expanding determinant through row \[{R_3}\],
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)\left| {\begin{array}{*{20}{c}}
  {ab + ac}&{{a^2}} \\
  {{b^2}}&{bc + ab}
\end{array}} \right|\]
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)\{ (b + c)(c + a) - ab\} \]
$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)(bc + ab + {c^2} + ac - ab)\]
$ \Rightarrow $L. H. S = \[2abc{(a + b + c)^3}\]= R. H. S
Hence Proved.

Note: Such problems are easy but require a lot of concentration while doing. If there is lack of
concentration, then the problem may not be solved. Also, properties of determinant are important to
solve problems but without them the problem can be solved but the process is very complicated and
tedious as it includes many terms. Make the determinant as simple as possible to easily expand it.
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