$u(x,y) = ax^3 + bx^2y + cxy^2 + dy^3$ is harmonic for some values of a,b,c and d . Find an analytic function f(z) with u as its real part. By taking the partial derivitives and second partialy derivative with respect to x and y I have found that b=-3d and c=-3a but I don't understand the last bit of the question or how to do it.
 A: Using what you have so far, write $$u(x,y) = ax^3 -3d x^2y -3a xy^2 + dy^3.$$
The Cauchy-Riemann equations give $$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 3ax^2 -6dxy-3ay^2$$
and $$ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}= 3dx^2+6axy -3dy^2.$$ 
The first equation then tells you that $$v(x,y) = 3ax^2 y-3dxy^2-ay^3 +F(x)$$ while the second gives $$ v(x,y) = dx^3+ 3ax^2y-3dxy^2 + G(y).$$
Putting these together yields $$v(x,y) = 3ax^2y-3dxy^2-ay^3+dx^3$$ up to a additive constant. 
Thus the function $$f(x+iy) = (ax^3 -3d x^2y -3a xy^2 + dy^3) + i(3ax^2y-3dxy^2-ay^3+dx^3) +K$$ is analytic with the required real part. For a nicer form, we can let $y=0$, which gives $$ f(x) = ax^3+idx^3+K.$$ Since that is the function represented on the real line, our function can be seen to be $$f(z) = (a+id)z^3+K.$$
A: $u(x,y)=ax^3 -3dx^2y-3axy^2+dy^3$ from $CR$  $equation $ you get $v(x,y)=3ax^2y -ay^3 -3dxy^2+dx^3 +k$ where k is a constant from this you can  get $f(z)$  
