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Question Number 118090 by bemath last updated on 15/Oct/20

$$\mathrm{find}\left|\begin{array}{c}\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{\mathrm{c}}\mathrm{c}\mathrm{c}\\ \mathrm{a}\frac{{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{a}}\mathrm{a}\\ \mathrm{b}\mathrm{b}\frac{{\mathrm{a}}^2+{\mathrm{c}}^2}{\mathrm{b}}\end{array}\right|=?$$

Commented by bemath last updated on 15/Oct/20

$$\mathrm{yes}..\mathrm{thank}\mathrm{you}\mathrm{all}$$

Answered by MJS_new last updated on 15/Oct/20

$$\mathrm{just}\mathrm{calculate}\mathrm{it}?!$$$$=4abc$$

Answered by som(math1967) last updated on 15/Oct/20

$$\frac{1}{\mathrm{abc}}\left|\begin{array}{ccc}{\mathrm{a}}^2+{\mathrm{b}}^2& {\mathrm{c}}^2& {\mathrm{c}}^2\\ {\mathrm{a}}^2& {\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2\\ {\mathrm{b}}^2& {\mathrm{b}}^2& {\mathrm{c}}^2+{\mathrm{a}}^2\end{array}\right|$$$${\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2-{\mathrm{R}}_3$$$$\frac{1}{\mathrm{abc}}\left|\begin{array}{ccc}0& -{2\mathrm{b}}^2& -{2\mathrm{a}}^2\\ {\mathrm{a}}^2& {\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2\\ {\mathrm{b}}^2& {\mathrm{b}}^2& {\mathrm{c}}^2+{\mathrm{a}}^2\end{array}\right|$$$$\frac{-2}{\mathrm{abc}}\left|\begin{array}{ccc}0& {\mathrm{b}}^2& {\mathrm{a}}^2\\ {\mathrm{a}}^2& {\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2\\ {\mathrm{b}}^2& {\mathrm{b}}^2& {\mathrm{c}}^2+{\mathrm{a}}^2\end{array}\right|$$$${\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1,{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1$$$$\frac{-2}{\mathrm{abc}}\left|\begin{array}{ccc}0& {\mathrm{b}}^2& {\mathrm{a}}^2\\ {\mathrm{a}}^2& {\mathrm{c}}^2& 0\\ {\mathrm{b}}^2& 0& {\mathrm{c}}^2\end{array}\right|$$$$\frac{-2}{\mathrm{abc}}[0-{\mathrm{b}}^2({\mathrm{a}}^2{\mathrm{c}}^2-0)+{\mathrm{a}}^2(0-{\mathrm{b}}^2{\mathrm{c}}^2)]$$$$\frac{-2\times -{2\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2}{\mathrm{abc}}=4\mathrm{abc}\mathrm{ans}$$$$$$

Answered by FelipeLz last updated on 15/Oct/20

$$\frac{a^2+b^2}{c}[\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2+c^4}{ab}-ab]-c[\frac{a^3+{ac}^2}{b}-ab]+c[ab-\frac{b^3-{bc}^2}{a}]$$$$\frac{a^2+b^2}{c}[\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2+c^4}{ab}-ab]\frac{ab}{ab}+c[2ab-\frac{b^3+{bc}^2}{a}-\frac{a^3+{ac}^2}{b}]\frac{ab}{ab}$$$$\frac{(a^2+b^2)c}{ab}[a^2+b^2+c^2]+\frac{c}{ab}[2a^2{b}^2-b^4-b^2{c}^2-a^4-a^2{c}^2]$$$$\frac{c}{ab}[(a^2+b^2)(a^2+b^2+c^2)-{(a^2-b^2)}^2-c^2(a^2+b^2)]$$$$\frac{c}{ab}[(a^2+b^2)(a^2+b^2)-{(a^2-b^2)}^2]$$$$\frac{c}{ab}[a^4+2a^2{b}^2+b^4-a^4+2a^2{b}^2-b^4]$$$$\frac{4a^2{b}^{2}c}{ab}=4abc$$

Answered by 1549442205PVT last updated on 15/Oct/20

$$\mathrm{By}\mathrm{rule}\mathrm{to}\mathrm{calculate}\mathrm{value}\mathrm{of}\mathrm{a}$$$$\mathrm{determinant}\mathrm{of}\mathrm{degree}3\mathrm{we}\mathrm{have}:$$$$\mathrm{\Delta }=\frac{({\mathrm{a}}^2+{\mathrm{b}}^2)({\mathrm{b}}^2+{\mathrm{c}}^2)({\mathrm{c}}^2+{\mathrm{a}}^2)}{\mathrm{abc}}+2\mathrm{abc}$$$$-\frac{\mathrm{bc}({\mathrm{b}}^2+{\mathrm{c}}^2)}{\mathrm{a}}-\frac{\mathrm{ac}({\mathrm{a}}^2+{\mathrm{c}}^2)}{\mathrm{b}}-\frac{\mathrm{ab}({\mathrm{a}}^2+{\mathrm{b}}^2)}{\mathrm{c}}$$$$=\frac{1}{\mathrm{abc}}[({\mathrm{a}}^2+{\mathrm{b}}^2)({\mathrm{b}}^2+{\mathrm{c}}^2)({\mathrm{c}}^2+{\mathrm{a}}^2)$$$$-{\mathrm{b}}^2{\mathrm{c}}^2({\mathrm{b}}^2+{\mathrm{c}}^2)-{\left(\mathrm{ca}\right)}^2({\mathrm{c}}^2+{\mathrm{a}}^2)$$$$-{\left(\mathrm{ab}\right)}^2({\mathrm{a}}^2+{\mathrm{b}}^2)]+2\mathrm{abc}$$$$\frac{1}{\mathrm{abc}}[{\mathrm{a}}^4({\mathrm{b}}^2+{\mathrm{c}}^2)+{\mathrm{b}}^4({\mathrm{c}}^2+{\mathrm{a}}^2)+{\mathrm{c}}^4({\mathrm{a}}^2+{\mathrm{b}}^2)$$$$+{2\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2-{\mathrm{b}}^4({\mathrm{c}}^2+{\mathrm{a}}^2)-{\mathrm{c}}^4({\mathrm{a}}^2+{\mathrm{b}}^2)$$$$-{\mathrm{a}}^4({\mathrm{b}}^2+{\mathrm{c}}^2)]+2\mathrm{abc}$$$$=\frac{1}{\mathrm{abc}}.2{\left(\mathrm{abc}\right)}^2+2\mathrm{abc}=4\mathrm{abc}$$

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