Question:
If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=26 \] 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:
\[ 9^2 = 26+2(ab+bc+ca) \]
\[ 81 = 26+2(ab+bc+ca) \]
\[ 55 = 2(ab+bc+ca) \]
\[ ab+bc+ca = \frac{55}{2} \]
Now,
\[ a^3+b^3+c^3-3abc \]
\[ = 9\left(26-\frac{55}{2}\right) \]
\[ = 9\left(\frac{52-55}{2}\right) \]
\[ = 9\left(\frac{-3}{2}\right) \]
\[ = -\frac{27}{2} \]