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प्रश्न
Prove that:
`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`
उत्तर
`[(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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