Math, asked by ujjwalkant12, 1 year ago

plz answer this question quickly

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Answered by Anonymous
1
Hi there!

Given a + b + c = 0

⇒ a³ + b³ + c³ = 3abc ----(i)

Consider,

L.H.S = (a²/bc) + (b²/ca) + (c²/ab)

(a³ + b³ + c³)/abc

3abc/abc = 3   [from Eqn. (i)] = R.H.S

Cheers!
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