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Question Number 122555 by ZiYangLee last updated on 18/Nov/20

$$provethat\left|\begin{array}{ccc}a& 1& a^2(b+c)\\ b& 1& b^2(c+a)\\ c& 1& c^2(a+b)\end{array}\right|=0$$

Commented by bramlexs22 last updated on 18/Nov/20

$$\Rightarrow a\left|\begin{array}{c}1b^{2}c+{ab}^2\\ 1{ac}^2+{bc}^2\end{array}\right|-1.\left|\begin{array}{c}b{b}^{2}c+{ab}^2\\ c{ac}^2+{bc}^2\end{array}\right|+(a^{2}b+a^{2}c)\left|\begin{array}{c}b1\\ c1\end{array}\right|$$$$$$

Answered by $@y@m last updated on 18/Nov/20

$$=\left|\begin{array}{ccc}a& 1& a^2(b+c)\\ b& 1& b^2(c+a)\\ c& 1& c^2(a+b)\end{array}\right|$$$$=\left|\begin{array}{ccc}a-b& 0& a^2(b+c)-b^2(c+a)\\ b-c& 0& b^2(c+a)-c^2(a+b)\\ c& 1& c^2(a+b)\end{array}\right|,{byR}_1\to R_1-R_2\mathrm{\&}{R}_2\to R_2-R_3$$$$=\left|\begin{array}{ccc}a-b& 0& ab(a-b)+c(a^2-b^2)\\ b-c& 0& bc(b-c)+a(b^2-c^2)\\ 1& 1& c^2(a+b)\end{array}\right|$$$$=(a-b)(b-c)\left|\begin{array}{ccc}1& 0& ab+c(a+b)\\ 1& 0& bc+a(b+c)\\ 1& -c& c^2(a+b)\end{array}\right|$$$$=0(\because R_1=R_2)$$

Commented by ZiYangLee last updated on 18/Nov/20

$$\mathrm{Wow}\mathrm{now}\mathrm{i}\mathrm{understand}!\mathrm{thank}\mathrm{sir}....$$

Answered by MJS_new last updated on 18/Nov/20

$${ac}^2(a+b)+a^{2}b(b+c)+b^{2}c(c+a)-$$$$-({ab}^2(c+a)+{bc}^2(a+b)+a^{2}c(b+c))=$$$$=a^2{c}^2+{abc}^2+a^2{b}^2+a^{2}bc+b^2{c}^2+{ab}^{2}c-$$$$-({ab}^{2}c+a^2{b}^2+{abc}^2+b^2{c}^2+a^{2}bc+a^2{c}^2)=0$$

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