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Question

If x+y+z=0, then prove that ∣ ∣xaybzcyczaxbzbxcya∣ ∣=xyz∣ ∣abccabbca∣ ∣

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Solution

Given x+y+z=0
To prove:
∣ ∣xaybzcyczaxbzbxcya∣ ∣=xyz∣ ∣abccabbca∣ ∣
LHS =∣ ∣xaybzcyczaxbzbxcya∣ ∣
=xa(za.yaxb.xc)yb(yc.yaxb.zb)+zc(yc.xcza.zb)=xa(a2yzx2bc)yb(y2acb2xz)+zc(c2xyz2ab)=xyza3x3abcy3abc+b3xyz+c3xyzz3abc=xyz(a3+b3+c3)abc(x3+y3+z3)=xyz(a3+b3+c3)abc(3xyz)
[x+y+z=0x3+y3+z3=3xyz]
=xyz(a3+b3+c33abc) .....(i)
Now, RHS =xyz∣ ∣abccabbca∣ ∣=xyz∣ ∣a+b+cbca+b+caba+b+cca∣ ∣ [C1C1+C2+C3]
=xyz(a+b+c)∣ ∣1bc1ab1ca∣ ∣ [taking (a+b+c) common from C1]
=xyz(a+b+c)∣ ∣0bcca0acba1ca∣ ∣
[R1R1R3andR2R2R3]
Expanding along C1
=xyz(a+b+c)[1(bc)(ba)(ac)(ca)]=xyz(a+b+c)(b2abbc+ac+a2+c22ac)=xyz(a+b+c)(a2+b2+c2abbcca)
=xyz(a3+b3+c33abc) .....(ii)
From Eqs. (i) and (ii),
LHS=RHS
∣ ∣xaybzcyczaxbzbxcya∣ ∣=xyz∣ ∣abccabbca∣ ∣
Hence proved.


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