Question

If a + b + c = 9 and ab + bc + ca = 26, find the value of $a^3+b^3+c^3-3abc$.

Answer 1

a + b + c = 9, ab + bc + ca = 26

Squaring, we get

$(a+b+c{)}^2=(9{)}^2{a}^2+b^2+c^2+2(ab+bc+ca)=81\Rightarrow a^2+b^2+c^2+2\times 26=81\Rightarrow a^2+b^2+c^2+52=81\therefore a^2+b^2+c^2=81-52=29Now,a^3+b^3+c^3-3abc=(a+b+c)\left[\right(a^2+b^2+c^2-(ab+bc+ca)\left)\right]=9[29-26]=9\times 3=27$