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Question Number 122555 by ZiYangLee last updated on 18/Nov/20
prove that determinant ((a,1,(a^2 (b+c))),(b,1,(b^2 (c+a))),(c,1,(c^2 (a+b))))=0
$${prove}\:{that}\begin{vmatrix}{{a}}&{\mathrm{1}}&{{a}^{\mathrm{2}} \left({b}+{c}\right)}\\{{b}}&{\mathrm{1}}&{{b}^{\mathrm{2}} \left({c}+{a}\right)}\\{{c}}&{\mathrm{1}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix}=\mathrm{0} \\ $$
Commented by bramlexs22 last updated on 18/Nov/20
⇒a determinant (((1 b^2 c+ab^2 )),((1 ac^2 +bc^2 )))−1. determinant (((b b^2 c+ab^2 )),((c ac^2 +bc^2 )))+(a^2 b+a^2 c) determinant (((b 1)),((c 1)))
$$\Rightarrow{a}\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:{b}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:\:\:{ac}^{\mathrm{2}} +{bc}^{\mathrm{2}} }\end{vmatrix}−\mathrm{1}.\begin{vmatrix}{{b}\:\:\:\:\:{b}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} }\\{{c}\:\:\:\:{ac}^{\mathrm{2}} +{bc}^{\mathrm{2}} }\end{vmatrix}+\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}\right)\begin{vmatrix}{{b}\:\:\:\mathrm{1}}\\{{c}\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$ \\ $$
Answered by $@y@m last updated on 18/Nov/20
= determinant ((a,1,(a^2 (b+c))),(b,1,(b^2 (c+a))),(c,1,(c^2 (a+b)))) = determinant (((a−b),0,(a^2 (b+c)−b^2 (c+a))),((b−c),0,(b^2 (c+a)−c^2 (a+b))),(c,1,(c^2 (a+b)))) , byR_1 →R_1 −R_(2 ) & R_2 →R_2 −R_3 = determinant (((a−b),0,(ab(a−b)+c(a^2 −b^2 ))),((b−c),0,(bc(b−c)+a(b^2 −c^2 ))),(1,1,(c^2 (a+b)))) =(a−b)(b−c) determinant ((1,0,(ab+c(a+b))),(1,0,(bc+a(b+c))),(1,(−c),(c^2 (a+b)))) =0 (∵R_1 =R_2 )
$$=\begin{vmatrix}{{a}}&{\mathrm{1}}&{{a}^{\mathrm{2}} \left({b}+{c}\right)}\\{{b}}&{\mathrm{1}}&{{b}^{\mathrm{2}} \left({c}+{a}\right)}\\{{c}}&{\mathrm{1}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix} \\ $$$$=\begin{vmatrix}{{a}−{b}}&{\mathrm{0}}&{{a}^{\mathrm{2}} \left({b}+{c}\right)−{b}^{\mathrm{2}} \left({c}+{a}\right)}\\{{b}−{c}}&{\mathrm{0}}&{{b}^{\mathrm{2}} \left({c}+{a}\right)−{c}^{\mathrm{2}} \left({a}+{b}\right)}\\{{c}}&{\mathrm{1}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix}\:,\:{byR}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −{R}_{\mathrm{2}\:} \&\:{R}_{\mathrm{2}} \rightarrow{R}_{\mathrm{2}} −{R}_{\mathrm{3}} \\ $$$$=\begin{vmatrix}{{a}−{b}}&{\mathrm{0}}&{{ab}\left({a}−{b}\right)+{c}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\\{{b}−{c}}&{\mathrm{0}}&{{bc}\left({b}−{c}\right)+{a}\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}\\{\mathrm{1}}&{\mathrm{1}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix}\:\: \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{\mathrm{1}}&{\mathrm{0}}&{{ab}+{c}\left({a}+{b}\right)}\\{\mathrm{1}}&{\mathrm{0}}&{{bc}+{a}\left({b}+{c}\right)}\\{\mathrm{1}}&{−{c}}&{{c}^{\mathrm{2}} \left({a}+{b}\right)}\end{vmatrix}\:\: \\ $$$$=\mathrm{0}\:\left(\because{R}_{\mathrm{1}} ={R}_{\mathrm{2}} \right) \\ $$
Commented by ZiYangLee last updated on 18/Nov/20
Wow now i understand ! thank sir....
$$\mathrm{Wow}\:\mathrm{now}\:\mathrm{i}\:\mathrm{understand}\:!\:\mathrm{thank}\:\mathrm{sir}…. \\ $$
Answered by MJS_new last updated on 18/Nov/20
ac^2 (a+b)+a^2 b(b+c)+b^2 c(c+a)− −(ab^2 (c+a)+bc^2 (a+b)+a^2 c(b+c))= =a^2 c^2 +abc^2 +a^2 b^2 +a^2 bc+b^2 c^2 +ab^2 c− −(ab^2 c+a^2 b^2 +abc^2 +b^2 c^2 +a^2 bc+a^2 c^2 )=0
$${ac}^{\mathrm{2}} \left({a}+{b}\right)+{a}^{\mathrm{2}} {b}\left({b}+{c}\right)+{b}^{\mathrm{2}} {c}\left({c}+{a}\right)− \\ $$$$\:\:\:\:\:−\left({ab}^{\mathrm{2}} \left({c}+{a}\right)+{bc}^{\mathrm{2}} \left({a}+{b}\right)+{a}^{\mathrm{2}} {c}\left({b}+{c}\right)\right)= \\ $$$$={a}^{\mathrm{2}} {c}^{\mathrm{2}} +{abc}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{ab}^{\mathrm{2}} {c}− \\ $$$$\:\:\:\:\:−\left({ab}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{abc}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$

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