Question

If a+b+c=10, a^2+b^2+c^2=54,a^3+b^3+c^3=352 then find the value of ab+bc+ca and abc

Answer 1

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
(10)^2=54+2(ab+bc+ca)
56/2=ab+bc+ca
28=ab+bc+ca

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca)
352-3abc=10(54-28)
352-3abc=260
92/3=abc

Answer 2

applying the formula of a²+b²+c² we get ( a+ b+ c) ²-2( ab+ bc + ca) so ab+ bc +ca= 23 and a³+ b³+ c³= ( a+ b+ c){ a²+b²+c²-(ab+bc+ca)}+3abc. then 3abc=352-230=>abc= 122/3.