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Question

Prove that ∣ ∣ ∣ab+ca2bc+ab2ca+bc2∣ ∣ ∣=(a+b+c)(ab)(bc)(ca)

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Solution

∣ ∣ ∣ab+ca2bc+ab2ca+bc2∣ ∣ ∣=∣ ∣ ∣a+b+cb+ca2a+b+cc+ab2a+b+ca+bc2∣ ∣ ∣C1C1+C2
Taking out common a+b+c from C1
=(a+b+c)∣ ∣ ∣1b+aa21c+ab21a+bc2∣ ∣ ∣=(a+b+c)∣ ∣ ∣1b+ca20abb2a20acc2a2∣ ∣ ∣R2R2R1R3R3R1=(a+b+c)∣ ∣ ∣1b+ca20(ab)(a+b)(ab)0(ac)(ac)(a+c)∣ ∣ ∣
Taking out common (ab) and (ac) from R1 and R2 respectively
=(a+b+c)(ab)(ac)∣ ∣ ∣1b+ca201(a+b)01(a+c)∣ ∣ ∣
Expanding along C1
=(a+b+c)(ab)(ac)[ac+a+b]
=(a+b+c)(ab)(ac)(bc)
=(a+b+c)(ab)(bc)(ca)
Proved

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