Question

If a+b+c =11 and (ab+bc+ca)=20,then find (se+b3+c3-3abc)=?

Answer 1

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Answer 2

a³ + b³ + c³ - 3abc = 671 if a+b+c =11 and (ab+bc+ca)=20

Step-by-step explanation:

Correct Question is

if If a+b+c =11 and (ab+bc+ca)=20

then find a³ + b³ + c³ - 3abc

a+b+c =11

Squaring both sides

a² + b² + c²  + 2ab + 2bc + 2ca  = 121

=> a² + b² + c² + 2 * 20 = 121

=> a² + b² + c² = 81

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

=> a³ + b³ + c³ - 3abc = 11 ( 81 - 20)

=>  a³ + b³ + c³ - 3abc = 11 * 61

=>  a³ + b³ + c³ - 3abc = 671

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