Question -13

Algebra Level 4

If x = c y + b z x=cy+bz , y = a z + c x y=az+cx and z = b x + a y z=bx+ay , where x , y , z ≠ 0 x,y,z \ne 0 , then find the value of ( a 2 + b 2 + c 2 ) (a^2+b^2+c^2) .

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1 − a b c 1-abc 1 + a b c 1+abc 1 − 2 a b c 1-2abc 1 + 2 a b c 1+2abc

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Karan Siwach
Mar 2, 2015

jee mains style :p

let x = y = z = 1 x=y=z=1

this will give us a + b = b + c = c + a = 1 a+b=b+c=c+a=1

and thus a = b = c = 1 / 2 a=b=c=1/2

and now just checkout for the right option !

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i did the same.....don't know why this question is level 4??? (As of now)

Kislay Raj - 6 years, 3 months ago

@Sandeep Bhardwaj Sir i cant publish a solution as i pressed wrong option :P .

For exact and short solution we can use cramer's rule . For having non trivial solution Delta = 0 . solving this we get a^2+b^2+c^2 = 1-2abc

Prakhar Bindal - 4 years, 3 months ago

someone pls provide legitimate solution. i tried multiplying equations to get ax + by + cz on LHS, and equated the modulus of the 2 sides of equation, but did not get answer in format.

Saket Joshi - 6 years, 3 months ago
Anubhav Bhatia
Mar 7, 2015

We have x=cy+bz now put value of y in this equation
we will get (1-c^2)X=(ac+b)Z .......1 Now put value of y in third equation
We will have (ac+b)X=(1-a^2)Z.........2 Divide 1. By 2 We will reach to req.d answer

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