The constant term in the expansion of ${\left( {\frac{x}{a} + \frac{{{a^2}}}{x}} \right)^6}$ , where $a \in \mathbb{R}$ is $1280$. Find $a$ .

evidence of binomial expansion     (M1)

eg selecting correct term,${\left( {\frac{x}{a}} \right)^6}{\left( {\frac{{{a^2}}}{x}} \right)^0} + \left( \begin{array}{l} 6\\ 1 \end{array} \right){\left( {\frac{x}{a}} \right)^5}{\left( {\frac{{{a^2}}}{x}} \right)^1} +  \ldots$

evidence of identifying constant term in expansion for power $6$     (A1)

eg   $r = 3$ , 4^th term

evidence of correct term (may be seen in equation)     A2

eg   $20\frac{{{a^6}}}{{{a^3}}}$ , $\left( \begin{array}{l} 6\\ 3 \end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3}$

attempt to set up their equation     (M1)

eg   $\left( \begin{array}{l} 6\\ 3 \end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3} = 1280$, ${a^3} = 1280$

correct equation in one variable $a$     (A1)

eg   $20{a^3} = 1280$ , ${a^3} = 64$

$a = 4$     A1     N4

[7 marks]

Many candidates struggled with this question. Some had difficulty with the binomial expansion, while others did not understand that the constant term had no $x$ , while still others were unable to simplify a ratio of exponentials with a common base. Some candidates found $r = 3$ using algebraic methods while others found it by writing out the first several terms. In some cases, candidates just set the entire expansion equal to 1280.