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Question Number 194642 by horsebrand11 last updated on 12/Jul/23

$$\mathrm{If}\mathrm{A}=\left(\begin{array}{c}\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{b}\mathrm{c}\mathrm{a}\\ \mathrm{c}\mathrm{a}\mathrm{b}\end{array}\right)\mathrm{and}\mathrm{a},\mathrm{b},\mathrm{c}>0$$$$\mathrm{such}\mathrm{that}\mathrm{abc}=1\mathrm{and}{\mathrm{A}}^{\mathrm{T}}.\mathrm{A}=\mathrm{I}$$$$\mathrm{find}{\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3-3\mathrm{abc}.$$

Answered by som(math1967) last updated on 12/Jul/23

$$A.A^T=I$$$$\left(\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right)\left(\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$$$$\left(\begin{array}{ccc}{a}^2+b^2+c^2& ab+bc+ca& ab+bc+ca\\ ab+bc+ca& a^2+b^2+c^2& ab+bc+ca\\ ab+bc+ca& ab+bc+ca& a^2+b^2+c^2\end{array}\right)$$$$=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$$$$\therefore a^2+b^2+c^2=1$$$$ab+bc+ca=0$$$${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$$$$(a+b+c)=\sqrt{1+0}=1[a,b,c>0]$$$$a^3+b^3+c^3-3abc$$$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$$$=1\times 1=1$$

Answered by cortano12 last updated on 12/Jul/23

$${\mathrm{A}}^{\mathrm{T}}\mathrm{A}=\mathrm{I}$$$$\mid \mathrm{A}{\mid }^2=1$$$$\mid \mathrm{A}\mid =\mathrm{a}(\mathrm{bc}-{\mathrm{a}}^2)-\mathrm{b}({\mathrm{b}}^2-\mathrm{ac})+\mathrm{c}(\mathrm{ab}-{\mathrm{c}}^2)=\pm 1$$$$3\mathrm{abc}-({\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3)=\pm 1$$$$\overline{)\begin{array}{c}{\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3-3\mathrm{abc}=\mp 1\end{array}}$$

Commented by manxsol last updated on 14/Jul/23

$$muybien$$

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