Factorize :
$125 a^{3} + b^{3} + 64 c^{3} - 60 a b c$

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Solution

$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Here $a = 5 a, b = b, c = 4 c$

On substitution: $= (5 a + b + 4 c) (25 a^{2} + b^{2} + 16 c^{2} - 5 a b - 4 b c - 20 a c)$

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125 a^{3} + b^{3} + 64 c^{3} - 60 a b c

125 a^{3} + b^{3} + 64 c^{3} - 60 a b c

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

125 a^{3} + \frac{1}{8}

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