Find the coefficient of \({x^{ - 2}}\) in the expansion of \({(x - 1)^3}{\left( {\frac{1}{x} + 2x} \right)^6}\).

expanding \({(x - 1)^3} = {x^3} - 3{x^2} + 3x - 1\)     A1

expanding \({\left( {\frac{1}{2} + 2x} \right)^6}\) gives

\(64{x^6} + 192{x^4} + 240{x^2} + \frac{{60}}{{{x^2}}} + \frac{{12}}{{{x^4}}} + \frac{1}{{{x^6}}} + 160\)     (M1)A1A1

Note:     Award (M1) for an attempt at expanding using binomial.

     Award A1 for \(\frac{{60}}{{{x^2}}}\).

     Award A1 for \(\frac{{12}}{{{x^4}}}\).

\(\frac{{60}}{{{x^2}}} \times  - 1 + \frac{{12}}{{{x^4}}} \times  - 3{x^2}\)     (M1)

Note:     Award (M1) only if both terms are considered.

therefore coefficient \({x^{ - 2}}\) is \( - 96\)     A1

Note:     Accept \( - 96{x^{ - 2}}\)

Note:     Award full marks if working with the required terms only without giving the entire expansion.

[6 marks]