if a+b+c=15 and a^2+b^2+c^2=83 find the value of a^3+b^3+c^3-3abc

Solution:-Given that a+b+c=15 and a^2+b^2+c^2=83

So,(a+b+c)^2=15^2

a^2+b^2+c^2+2(ab+bc+ca)=225

83+2(ab+bc+ca)=225+

2(ab+bc+ca)=225-83

2(ab+bc+ca)=142

(ab+bc+ca)=142/2

(ab+bc+ca)=71

Now,we know that 

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca)) 

Put in a^3+b^3+c^3-3abc=15×(83-71) 

a^3+b^3+c^3-3abc=15×12

a^3+b^3+c^3-3abc=180 ans....