Question

a+b+c=6,ab+bc+ca=11,thena^3+b^3+c^3-3abc=?

Answer 1

Hi friend
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Your answer
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Given that : - (a + b + c) = 6 and (ab + bc + ca ) = 11.

To calculate : - Value of (a³ + b³ + c³ - 3abc).

Now,
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(a + b + c)² = (a² + b² + c² - ab - bc - ca)

=> (6)² = [ (a² + b² + c²) - (ab + bc + ca) ]

=> 36 = (a² + b² + c²) - 11

=> (a² + b² + c²) = 36 + 11

=> (a² + b² + c²) = 47

Then,
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a³ + b³ + c³ - 3abc

=> (a + b + c)(a² + b² + c² - ab - bc - ca)

=> 6 × [(a² + b² + c²) - (ab + bc + ca)]

=> 6 × (47 - 11)

=> 6 × 36

=> 216

HOPE IT HELPS