Ask a Teacher

If a+b+c=12, a^2+b^2+c^2=90, Find the value of a^3+b^3+c^3-3abc.

a+b+c = 12
Squaring both the sides we get

(a+b+c)² = (12)²
( (a² + b² +c² )+2ab +2bc +2ac )= 144
( 90 + 2(ab + bc + ac ) ) = 144
2(ab+bc+ac) = 144-90
= 54
ab+bc+ac = 54 / 2
= 27

a³+b³+c³-3abc = (a+b+c)(a²+ b²+ c² -ab -bc -ac)
= 12 ( 90 - ( ab +bc + ac )
= 12 ( 90 - 27)
= 12 * 63
= 756