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Question

If p+q+r=0, then prove that

∣ ∣paqbrcqcrapbrbpcqa∣ ∣=pqr∣ ∣abccabbca∣ ∣.

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Solution


Given p+q+r=0

p3+q3+r3=3pqr

p3+q3+r23pqr=0 (i)

R.H.S=pqr∣ ∣abccabbca∣ ∣

Applying C1C1+C2+C3

=pqr∣ ∣a+b+cbca+b+caba+b+cca∣ ∣

=pqr(a+b+c)∣ ∣1bc1ab1ca∣ ∣

Applying R2R2R1 and R3R3R1

=pqr(a+b+c)∣ ∣1bc0abbc0cbac∣ ∣

=pqr(a+b+c)abbccbac

=pqr(a+b+c)(a2+b2+c2abbcca)

=pqr(a3+b2+c22abc)

=pqr(a3+a3+c33abc)abc(p3+q3+r33pqr) .......{from (i) }

=pqr(a3+b3+c3)abc(p3+q3+r3)

=(pqra3abcp3)(abcq3pqrb3)+(pqrc3abcr3)

=(ap){(ra)(qa)(pb)(pc)}(qb){(pc)(qa)(pb)(rb)}+(rc){(pc)(qc)(ra)(rb)}

=∣ ∣paqbrcqcrapbrbpcqa∣ ∣=L.H.S.
Hence proved

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