if a+b+c=12; a2+b2+c2=90,find the value of a3+b3+c3-3abc
PLEASE ANSWER FAST AS I HAVE MY MATHS TEST TOMMOROW
It is given that a+b+c = 12
Squaring both the sides we get
(a+b+c)^2 = (12)^2
a^2 + b^2 +c^2 + 2ab +2bc +2ac = 144
90 + 2(ab + bc + ac ) = 144
2(ab+bc+ac) = 144-90 = 54
ab+bc+ac = 54 / 2 = 27
Now use this identity, it is given in your book.
a^3+b^3+c^3-3abc = (a+b+c)(a^2+ b^2+ c^2 -ab -bc -ac)
= 12 [90 - ( ab +bc + ac ) ]
= 12 ( 90 - 27)
= 12 * 63
= 756