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Using properties of determinants, prove the following $∣ ∣ 3 a a - b a - c - a + b 3 b b - c - a + c c - a 3 c ∣ ∣ = 3 (a + b + c) (a b + b c + c a)$

Name: Video solution 1: Using properties of determinants, prove the following ∣∣3aa−ba−c−a+b3bb−c−a+cc−a3c∣∣=3(a+b+c)(ab+bc+ca)
Uploaded: 2021-01-01T18:41:02.000Z
Duration: 2 min
Description: Video solution 1: Using properties of determinants, prove the following ∣∣3aa−ba−c−a+b3bb−c−a+cc−a3c∣∣=3(a+b+c)(ab+bc+ca)

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Using properties of determinants, prove the following

∣ ∣ 3 a a - b a - c - a + b 3 b b - c - a + c c - a 3 c ∣ ∣ = 3 (a + b + c) (a b + b c + c a)

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Question Text	Using properties of determinants, prove the following $∣ ∣ 3 a a - b a - c - a + b 3 b b - c - a + c c - a 3 c ∣ ∣ = 3 (a + b + c) (a b + b c + c a)$
Answer Type	Video solution: 1
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Using properties of determinants, prove the following ∣∣​3aa−ba−c​−a+b3bb−c​−a+cc−a3c​∣∣​=3(a+b+c)(ab+bc+ca)

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Using properties of determinants, prove the following $∣ ∣ 3 a a - b a - c - a + b 3 b b - c - a + c c - a 3 c ∣ ∣ = 3 (a + b + c) (a b + b c + c a)$