I just wanted to see if I could get Alan's answer using a different (albeit, more complicated) approach......
So we have
C(n,3)(3x)^3 + C(n,2)(3x)^2
C(n,3)(27x^3) + C(n,2)(9x^2)
27C(n,3)x^3 + 9C(n,2)x^2 and we're told that
27C(n,3) = (10)[(9)C(n,2)] divide by 9 on both sides
3C(n,3) = 10C(n,2)
3[n!/((n-3)!3!)] = 10[n!/((n-2)!2!)] divide by n! on both sides
3/[(n-3)!3!)] = 10/[(n-2)!2!] simplify
(1/2)/(n-3)! = 5/(n-2)! multiply by 2 on each side
1/(n-3)! = 10/(n-2)! multiply both sides by (n-2)!
(n-2)!/(n-3)! = 10 the left side simplifies to (n-2)
n-2 = 10
So....n = 12......just as Alan found !!!
