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Question

Use proportion of determinants to solve
∣ ∣ ∣aa2bcbb2cacc2ab∣ ∣ ∣=(ab)(bc)(ca)(ab+bc+ca)

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Solution

Consider the following question

∣ ∣ ∣aa2bcbb2cacc2ab∣ ∣ ∣=(ab)(bc)(ca)(ab+bc+ca)

Applying R2R2R1 R3R3R1

1abc∣ ∣ ∣a2a3abcb2b3bcac2c3cab∣ ∣ ∣=abcabc∣ ∣ ∣a2a31b2b31c2c31∣ ∣ ∣

∣ ∣ ∣a2a31b2b31c2c31∣ ∣ ∣=∣ ∣ ∣a2a31b2a2b3a30c2a2c3a30∣ ∣ ∣

=(b2a2)(c3a3)(b3a3)(c2a2)

=(bc)(b+c)(ca)(c2+a2+ac)(b2+a2+ab)(a+b)(ca)(c+a)

=(ca)(ba)((bc)(b+c)(c2+a2+ac)(b2+a2+ab)(a+b)(c+a))

=(ca)(ba)(ac)(ab+bc+ca)


Hence, this is the required answer.


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