Question

Find a^3+b^3+c^3-3abc ifa+b+c=5 and a^2+b^2+c^2= 29

Answer 1

We know that ,

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

Now,

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

5^2=29+2(ab+bc+ca)

25-29=2(ab+bc+ca)

-4=2(ab+bc+ca)

ab+bc+ca=-4/2

ab+bc+ca=-2 --------------------------------------------------eq.1


Now,

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=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca)] -------------eq.2

By putting the value of eq.1 in eq.2,

a^3+b^3+c^3-3abc=(5)[29-(-2)]

a^3+b^3+c^3-3abc=5(29+2)

a^3+b^3+c^3-3abc=5*31

a^3+b^3+c^3-3abc=155