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

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

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

Answered by A5T last updated on 26/Jan/25

x=bz+cy...(i); y=cx+az...(ii); z=bx+ay...(iii) (i) in (ii)⇒ y=cbz+c^2 y+az...(iv) (i) in (iii)⇒ z=b^2 z+bcy+ay⇒z=((bcy+ay)/(1−b^2 ))...(v) (v) in (iv) ⇒y=((b^2 c^2 y+2abcy+a^2 y)/(1−b^2 ))+c^2 y...(vi) (vi)/y⇒1=((b^2 c^2 +2abc+a^2 )/(1−b^2 ))+c^2 =((2abc+a^2 +c^2 )/(1−b^2 )) ⇒1−b^2 =a^2 +c^2 +2abc⇒a^2 +b^2 +c^2 +2abc=1 ■

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

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

−x+cy+bz=0 cx−y+az=0 bx+ay−z=0 such that nontrivial solution exists, determinant (((−1),c,b),(c,(−1),a),(b,a,(−1)))=0 −(1−a^2 )−c(−c−ab)+b(ca+b)=0 ⇒a^2 +b^2 +c^2 +2abc=1 ✓

$$−{x}+{cy}+{bz}=\mathrm{0} \\ $$$${cx}−{y}+{az}=\mathrm{0} \\ $$$${bx}+{ay}−{z}=\mathrm{0} \\ $$$${such}\:{that}\:{nontrivial}\:{solution}\:{exists}, \\ $$$$\begin{vmatrix}{−\mathrm{1}}&{{c}}&{{b}}\\{{c}}&{−\mathrm{1}}&{{a}}\\{{b}}&{{a}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$−\left(\mathrm{1}−{a}^{\mathrm{2}} \right)−{c}\left(−{c}−{ab}\right)+{b}\left({ca}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{abc}=\mathrm{1}\:\checkmark \\ $$

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