Question

pls solve these 2 or any one of these​

Answer 1

Answer =

We know that (a + b + c)³ = (a³ + b³ + c³) + 3[(a + b + c)(ab +

ac + bc) - abc]

If (a + b + c) = 0, then a³ + b³ + c³ = 3abc

Given, 3a + 5b + 4c = 0

Here, a = 3a, b = 5b and c = 4c

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Now,

L.H.S = 27a³ + 125b³ + 64c³

= (3a)³ + (5b)³ + (4c)³

=

= 3(3a)(5b)(4c)

= 180abc

=R.H.S