Verify that $27 a^{3} + b^{3} + c^{3} - 9 a b c = (3 a + b + c) [9 a^{2} + b^{2} + c^{2} - 3 a b - b c - 3 a c]$

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Solution

L.H.S. = $27 a^{3} + b^{3} + c^{3} - 9 a b c$
$= (3 a)^{3} + (b)^{3} + (c)^{3} - 3 (3 a) (b) (c)$ $∵ [a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)]$
$= (3 a + b + c) [(3 a)^{2} + (b)^{2} + (c)^{2} - 3. a . b - b . c - c .3 a]$
$= (3 a + b + c) [9 a^{2} + b^{2} + c^{2} - 3 a b - b c - 3 a c]$ Proved

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