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Question

By using properties of determination, show that
∣ ∣abc2a2a2bbca2b2c2ccab∣ ∣=(a+b+c)3

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Solution

R1R1+R2+R3
∣ ∣a+b+ca+b+ca+b+c2bbca2b2c2ccab∣ ∣=(a+b+c)∣ ∣1112bbca2b2c2ccab∣ ∣
c2c2c1; c3c3c1
=(a+b+c)∣ ∣1112babc02c2cabc∣ ∣
=(a+b+c)(a+b+c)(a+b+c)=(a+b+c)3

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