Question

6. If a + b + c = 15 and a2

+ b2

+ c2

= 83, find the value of a3

+ b3

+ c3

– 3abc.

Answer 1

Hey dear,

Here is ur answer,

$( {a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

${15}^{2} = 83 + 2(ab + bc + ca)$

ab+bc+ca=( 225-83)/2 => 71.

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

$= > \: 15(83 - 71)$

= 15×12

=180.