Question

If a+b+c=15 and a^2+b^2+c^2=68 find the value of a^3+b^3+c^3-3abc

Answer 1

a² + b² + c² = 68
a + b + c = 15

use the concept ,
(a +b + c)² = a² + b² + c² +2(ab+bc+ca)
(15)²= 68 + 2( ab + bc + ca)
225 -68 = 2(ab + bc +ca)
157/2 = ab + bc + ca

ab + bc + ca = 78.5

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

=( 15)( 68 -78.5)
=(15)(-10.5)
= -157.5 ( answer )

Answer 2

since a+ b+ c = 15 then a²+b²+c²=( a+ b+ c) ²- 2(ab+ bc+ ca) , so ab+ bc +ca = 157/2.and a³+b³+c³=( a+ b+ c)( a²+b²+c²-ab- bc- ca) =15.( 225-157/2)= 2197.5