# 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

363.9k+ views

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.

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.

Last updated date: 01st Oct 2023

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Total views: 363.9k

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