Question

Of a+b+c=5,ab+bc+ca=10 prove a cube +b cube + c cube - 3 abc =-25

Answer 1

Answer:

a³+b³+c³-3abc=

(a+b+c)(a²+b²+c²-ab-bc-ca)

a+b+c=5 ,ab+bc+ca=10 ,a²+b²+c²=?

a²+b²+c²=(a+b+c)²-2ab-2bc-2ca

=(a+b+c)²-2(ab+bc+ca)

=(5)²-2(10)

=25-20

=5

a³+b³+c³-3abc=

(a+b+c)(a²+b²+c²-ab-bc-ca)

=(a+b+c)[(a²+b²+c²)-(ab+bc+ca)]

=(5)[(5)-(10)]

=(5)(5-10)

=5×(-5)

=-25

•°•a³+b³+c³-3abc=-25