# Find the coefficient of a term in the expansion of an algebraic expression

The coefficient of $$x^3$$ in the expansion of (1 + 2$$x$$ + 3$$x^2$$)$$^6$$ is equal to twice the coefficient of $$x^4$$ in the expansion of $$(1 - a x^2)^5$$.

Find all possible value of constant $$a$$.

I am actually getting $$a=8$$ or $$a=-8$$ using the binomial theorem or Pascal's Triangle.

Hint $$:$$ Observe that \begin{align} (1+2x+3x^2)^6 & = 1 + \binom 6 1 (2x+3x^2) + \binom 6 2 (2x+3x^2)^2 + \binom 6 3 (2x+3x^2)^3 + \cdots + (2x+3x^2)^6. \\ & = 1 + \binom 6 1 x(2+3x) + \binom 6 2 x^2(2+3x)^2 + \binom 6 3 x^3(2+3x)^3 + \cdots + x^6(2+3x)^6. \end{align} Can you see now?
So the coefficient of $$x^3$$ in $$(1+2x+3x^2)^6$$ is $$\left (12 \times \binom 6 2 \right ) + \left (8 \times \binom 6 3 \right ) = 180 + 160 = 340.$$
Now the coefficient of $$x^4$$ in $$(1-ax^2)^5$$ is $$\binom 5 2 a^2 = 10 a^2.$$
So according to the given problem we must have $$2 \times 10a^2 = 340 \implies a^2 = 17.$$ Therefore $$a = \pm {\sqrt {17}}.$$