27(a $-$ b)³ + (2a $-$ b)³ + (4b $-$ 5a)³

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Solution

$We have 27 {(a - b)}^{3} + {(2 a - b)}^{3} + {(4 b - 5 a)}^{3} = {[3 (a - b)]}^{3} + {(2 a - b)}^{3} + {(4 b - 5 a)}^{3} Let 3 (a - b) = x, 2 a - b = y and 4 b - 5 a = z ∴ {[3 (a - b)]}^{3} + {(2 a - b)}^{3} + {(4 b - 5 a)}^{3} = x^{3} + y^{3} + z^{3} Here, x + y + z = 3 (a - b) + (2 a - b) + (4 b - 5 a) = 0 However, we know that if x + y + z = 0, then x^{3} + y^{3} + z^{3} = 3 x y z ∴ {[3 (a - b)]}^{3} + {(2 a - b)}^{3} + {(4 b - 5 a)}^{3} = 3 [3 (a - b)] (2 a - b) (4 b - 5 a)$

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