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If x=bz+CY;y=az+cz; z=ay+bx then proved a²+b²+c²+2abc=1

If x=bz+CY;y=az+cz; z=ay+bx then proved a²+b²+c²+2abc=1

Grade:10

1 Answers

Arun
25750 Points
6 years ago
Dear satyajit
 
Since, x=cy+bz----(1)
y=az+cx----(2)
and z=bx+ay----(3)
by cancellation z from (1)&(2) by help (3) we get,
(1-b^2)x=(c+ab) y
=>x/y=(c+ab)/(1-b^2) ----(4)
&(1-a^2)y=(c+ab) x
=>y/x=(c+ab)/(1-a^2) -----(5)
now by multiplying(4)&(5) we get,
1=(c^2+2abc+a^2b^2)/(1-a^2-b^2+a^2b^2)
=> a^2+b^2+c^2+2abc=1
 
Regards
Arun (askIITIANS FORUM EXPERT)

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