Question:
If
\[ a+b+c=9 \] and \[ ab+bc+ca=23, \] then \[ a^3+b^3+c^3-3abc= \]
(a) 108
(b) 207
(c) 669
(d) 729
Solution:
Using identity:
\[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]
Now,
\[ a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) \]
Substituting the given values:
\[ a^2+b^2+c^2 = 9^2-2(23) \]
\[ =81-46 \]
\[ =35 \]
Therefore,
\[ a^2+b^2+c^2-ab-bc-ca = 35-23 \]
\[ =12 \]
Now,
\[ a^3+b^3+c^3-3abc = 9 \times 12 \]
\[ =108 \]
Hence, the correct answer is:
\[ \boxed{108} \]