F.4 ~ Binomial Theorem

Let Ar be the coefficient of x^2 in the expansion of [a(x^2) – (1 / 2x)]^6, where a ≠ 0. If 9A3 = -2A6, find the value of a.
[a(x^2) – (1 / 2x)]^6
=(ax^2-1/2x)^6
= a^6 x^12 - 6*a^5 x^10(1/2x) + 15*a^4 x^8(1/4x^2) - 20*(a^3 x^6)(1/8x^3)+15*a^2 x^4(1/16x^4)-6*ax^2(1/32x^5)+(1/64x^6)
=a^6 x^12 - 3a^5 x^9 + 15/4*a^4 x^6 - 5/2*a^3 x^3+15/16*a^2-3/16*a/x^3+(1/64x^6)
9*5/2*a^3=-2*5/4*a^4
45a^3/2=-10a^4/4
180a^3=-20a^4
a=-9