If a + b + c = 6 and a2 + b2 + c2 = 40 then, a3 + b3 + c3 - 3abc = ?

A

412

B

232

C

180

D

252

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Question: If a + b + c = 6 and a2 + b2 + c2 = 40 then, a3 + b3 + c3 - 3abc = ?

​Solution:

Given that,

​ a + b + c = 6

​a² + b² + c² = 40

​Now,

​a + b + c = 6

(a + b + c)2 = 62

a2 + b2 + c2 + 2ab + 2bc + 2ac = 36

40 + 2(ab + bc + ca) = 36

2(ab + bc + ca) = - 4

ab + bc + ca = - 2

​Then,

​a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)

= 6[40 - (-2)]

​= 6[40 + 2]

​= 6 × 42

​= 252

​a3 + b3 + c3 - 3abc = 252

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Updated: 12 hours ago

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