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Question

Using properties of determinants, prove the following:
∣ ∣abcabbccab+cc+aa+b∣ ∣=a3+b3+c33abc

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Solution

Let us take the LHS of the given expression.
LHS=∣ ∣abcabbccab+cc+aa+b∣ ∣

Perform C1C1+C2+C3LHS=∣ ∣a+b+cbc0bcca2(a+b+c)c+aa+b∣ ∣

Take (a+b+c) common from C1,
LHS=(a+b+c)∣ ∣1bc0bcca2c+aa+b∣ ∣

Perform R3R32R1LHS=(a+b+c)∣ ∣1bc0bcca0c+a2ba+b2c∣ ∣

Now, expanding the determinant along C1, we get
LHS=(a+b+c)((bc)(a+b2c)(c+a2b)(ca))
=(a+b+c)(ab+b22bcacbc+2c2c2+acac+a2+2bc2ab)
=(a+b+c)(a2+b2+c2abbcac)

We know that
a3+b3+c3=(a+b+c)(a2+b2+c2abbcac)+3abc
a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcac)
LHS=a3+b3+c33abc

Thus, LHS = RHS

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