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Prove that: [abca-bb-cc-ab+cc+aa+b]=a3+b3+c3-3abc - Mathematics

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Question

Prove that:

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`

Sum

Solution

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)]`

L.H.S = `[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)]`

Apply C1 → C1 + C2 + C3

= `[(a + b + c, b, c),(0, b - c, c - a),(2(a + b + c), c + a, a + b)]`

Taking (a + b + c) common from C1 we get,

= (a + b + c) `|(1, b, c),(0, b - c, c - a),(2, c + a, a + b)|`

Applying, R3 → R3 − 2R1

= `(a + b + c) |(1, b, c),(0, b - c, c - a),(0, c + a - 2b, a + b - 2c)|`

= (a + b + c)[(b − c)(a + b − 2c) − (c − a)(c + a − 2b)]

= a3 + b3 + c3 − 3abc

As, L.H.S = R.H.S

Hence, proved.

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Chapter 6: Determinants - Exercise 6.2 [Page 58]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 11 | Page 58

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