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]