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Question Number 216046 by MATHEMATICSAM last updated on 26/Jan/25

$$x=bz+cy,y=cx+az\mathrm{and}z=bx+ay$$$$\mathrm{then}\mathrm{prove}\mathrm{that}{a}^2+b^2+c^2+2abc=1.$$

Answered by A5T last updated on 26/Jan/25

$$\mathrm{x}=\mathrm{bz}+\mathrm{cy}...\left(\mathrm{i}\right);\mathrm{y}=\mathrm{cx}+\mathrm{az}...\left(\mathrm{ii}\right);\mathrm{z}=\mathrm{bx}+\mathrm{ay}...\left(\mathrm{iii}\right)$$$$\left(\mathrm{i}\right)\mathrm{in}\left(\mathrm{ii}\right)\Rightarrow \mathrm{y}=\mathrm{cbz}+{\mathrm{c}}^2\mathrm{y}+\mathrm{az}...\left(\mathrm{iv}\right)$$$$\left(\mathrm{i}\right)\mathrm{in}\left(\mathrm{iii}\right)\Rightarrow \mathrm{z}={\mathrm{b}}^2\mathrm{z}+\mathrm{bcy}+\mathrm{ay}\Rightarrow \mathrm{z}=\frac{\mathrm{bcy}+\mathrm{ay}}{1-{\mathrm{b}}^2}...\left(\mathrm{v}\right)$$$$\left(\mathrm{v}\right)\mathrm{in}\left(\mathrm{iv}\right)\Rightarrow \mathrm{y}=\frac{{\mathrm{b}}^2{\mathrm{c}}^2\mathrm{y}+2\mathrm{abcy}+{\mathrm{a}}^2\mathrm{y}}{1-{\mathrm{b}}^2}+{\mathrm{c}}^2\mathrm{y}...\left(\mathrm{vi}\right)$$$$\left(\mathrm{vi}\right)/\mathrm{y}\Rightarrow 1=\frac{{\mathrm{b}}^2{\mathrm{c}}^2+2\mathrm{abc}+{\mathrm{a}}^2}{1-{\mathrm{b}}^2}+{\mathrm{c}}^2=\frac{2\mathrm{abc}+{\mathrm{a}}^2+{\mathrm{c}}^2}{1-{\mathrm{b}}^2}$$$$\Rightarrow 1-{\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{c}}^2+2\mathrm{abc}\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+2\mathrm{abc}=1◼$$

Answered by mr W last updated on 26/Jan/25

$$-x+cy+bz=0$$$$cx-y+az=0$$$$bx+ay-z=0$$$$suchthatnontrivialsolutionexists,$$$$\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$$$$-(1-a^2)-c(-c-ab)+b(ca+b)=0$$$$\Rightarrow a^2+b^2+c^2+2abc=1✓$$

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