Question

if a+b+c=11 and ab +bc+ca=20 ,then (a^3+b^3+c^3-3abc)=??

Answer 1

given,
a+b+c=11
ab+bc+ca=20

as we know that,
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca
on putting values
11^2=a^2+b^2+c^2+ 2(ab+bc+ca)
121=a^2+b^2+c^2+2*20
121=a^2+b^2+c^2+40
121-40=a^2+b^2+c^2
81=a^2+b^2+c^2
        or
a^2+b^2+c^2=81

as we know that
a^3+b^3+c^3- 3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
a^3+b^3+c^3- 3abc =(11)(81-( ab +bc+ca))
a^3+b^3+c^3- 3abc=(11)(81-( 20))
a^3+b^3+c^3- 3abc=(11)(81-20)
a^3+b^3+c^3- 3abc=(11)(61)
a^3+b^3+c^3- 3abc=(671)