Residue of $\frac{z^3}{(z-1)(z-2)(z-3)}$ at $z=\infty$ Residue of $\dfrac{z^3}{(z-1)(z-2)(z-3)}$ at $z=\infty$. 
Well, the total sum of residues is zero. But the given answer says 6.
We calculate residues with $Res\;f(a)=\dfrac{1}{(n-1)!}\left[\dfrac{d^{n-1}}{dz^{n-1}}[(z-a)^nf(z)]\right]_{z=a}$, but that can't be used here ..
I don't know how to proceed. Please help.
Thanks.
 A: $$\operatorname{Res} \left[ {f\left( z \right),z = 1} \right] = \mathop {\lim }\limits_{z \to 1} \left( {z - 1} \right)f\left( z \right) = \mathop {\lim }\limits_{z \to 1} \frac{{{z^3}}}
{{\left( {z - 2} \right)\left( {z - 3} \right)}} = \frac{1}
{2},$$
$$\operatorname{Res} \left[ {f\left( z \right),z = 2} \right] = \mathop {\lim }\limits_{z \to 2} \left( {z - 2} \right)f\left( z \right) = \mathop {\lim }\limits_{z \to 2} \frac{{{z^3}}}
{{\left( {z - 1} \right)\left( {z - 3} \right)}} =  - 8,$$
$$\operatorname{Res} \left[ {f\left( z \right),z = 3} \right] = \mathop {\lim }\limits_{z \to 3} \left( {z - 3} \right)f\left( z \right) = \mathop {\lim }\limits_{z \to 3} \frac{{{z^3}}}
{{\left( {z - 1} \right)\left( {z - 2} \right)}} = \frac{{27}}
{2}$$
We have
$$\operatorname{Res} \left[ {f\left( z \right),z = 1} \right] + \operatorname{Res} \left[ {f\left( z \right),z = 2} \right] + \operatorname{Res} \left[ {f\left( z \right),z = 3} \right] + \operatorname{Res} \left[ {f\left( z \right),z = \infty } \right] = 0.$$
So
$$\operatorname{Res} \left[ {f\left( z \right),z = \infty } \right] =  - \left( {\operatorname{Res} \left[ {f\left( z \right),z = 1} \right] + \operatorname{Res} \left[ {f\left( z \right),z = 2} \right] + \operatorname{Res} \left[ {f\left( z \right),z = 3} \right]} \right) = -6$$
