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Question

Prove that:
∣ ∣ ∣(b+c)2abcaab(c+a)2bccabc(a+b)2∣ ∣ ∣=2abc(a+b+c)3

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Solution

L:H:S=∣ ∣ ∣(b+c)2abcaab(c+a)2bccabc(a+b)2∣ ∣ ∣

=1abc∣ ∣ ∣a(b+c)2a2ba2cab2b(c+a)2b2cac2bc2c(a+b)2∣ ∣ ∣

=∣ ∣ ∣(b+c)2a2a2b2(c+a)2b2c2c2(a+b)2∣ ∣ ∣

R1+C1C3,C2C2C3

=∣ ∣ ∣(b+ca)(b+c+a)0a20(c+a+b)(c+ab)b2(cab)(c+a+b)(cab)(c+a+b)(a+b)2∣ ∣ ∣

(a+b+c)2∣ ∣ ∣b+ca0a20c+abb2cabcab(a+b)2∣ ∣ ∣
R3+R3(R1+R2)

(a+b+c)2∣ ∣ ∣b+ca0a20c+abb22b2a2ab∣ ∣ ∣

(a+b+c)2ab∣ ∣ ∣a(b+ca)0a20b(c+ab)b22ab2ab2ab∣ ∣ ∣

(a+b+c)2ab∣ ∣ ∣a+b+ca2a2b2bc+abb2UU2ab∣ ∣ ∣

2abc(a+b+c)3
R:H:S (Proved)


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