Prove that|a2 bc ac+c2 a2+ab b2 ac ab b2+bc c2|=4a2b2c2
Solution and Explanation
Δ=|a2 bc ac+c2 a2+ab b2 ac ab b2+bc c2|
Taking out common factors a,b and c from C1,C2 and C3 we have,
Δ=abc|a c a+c a+b b a b b+c c|
Applying R2→R2-R1 and R3→R3-R1,we have:
Δ=abc|a c a+c b b-c -c b-a b -a |
Applying R2→R2+R1,we heve:
Δ=abc|a c a+c a+b b a b-a b -a|
Applying R3→R3+R2,we heve:
Δ=abc|a c a+c a+b b a 2b 2b 0|
=2a2bc|a c a+c a+b b a 2b 2b 0|
ApplyingC2→C2-C1,we heve:
Δ=2a2bc|a c-a a+c a+b -a a 100|
Expanding along R3,we have:
Δ=2a2bc[a(c-a)+a(a+c)]
=2a2bc[ac-a2+a2+ac]
=2a2bc(2ac)
=4a2b2c2
Hence,the given result is proved.
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