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If x=bz+CY;y=az+cz; z=ay+bx then proved a²+b²+c²+2abc=1
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 RegardsArun (askIITIANS FORUM EXPERT)
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