Question:
If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=35 \] find:
\[ a^3+b^3+c^3-3abc \]
Solution:
Using identity:
\[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]
First find \[ ab+bc+ca \]
Using identity:
\[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \]
Substituting the given values:
\[ 9^2 = 35+2(ab+bc+ca) \]
\[ 81 = 35+2(ab+bc+ca) \]
\[ 46 = 2(ab+bc+ca) \]
\[ ab+bc+ca = 23 \]
Now,
\[ a^3+b^3+c^3-3abc \]
\[ = 9(35-23) \]
\[ = 9\times12 \]
\[ =108 \]