$\displaystyle Since,\ f(z)=\ u+iv\ be\ analytic\ \\[2ex] \displaystyle By\ Cauchy\ rehmans\ equations \\[2ex] \displaystyle u_x=v_y\ and\ u_y=-v_x\ \ \ ……(A) \\[2ex] \displaystyle u-v=\left(x-y\right)\left(x^2+4xy+y^2\right)\\[2ex] u-v=x^3+4x^2y+xy^2-x^2y-4xy^2-y^3 \\[2ex] \displaystyle u-v=x^3+3x^2y-3xy^2-y^3 \\[2ex] $

$\displaystyle Differentiating\ partially\ \ w.r.t\ x,\ \ \ \\[2ex] \displaystyle u_x-v_x=3x^2+6xy-3y^2……(B) \\[2ex] \displaystyle Differentiating\ \ partially\ w.r.t\ \ y, \\[2ex] \displaystyle u_y-v_y=3x^2-6xy-3y^2……(C) \\[2ex] $

$From (A) and (B)\\ \displaystyle v_y-v_x=\ 3x^2+6xy-3y^2 \\[2ex] From (A) and (C)\\[2ex] \displaystyle -v_x-v_y=3x^2-6xy-3y^2\ \\[2ex] $

$Adding \ the \ above \ two \ equations,\\[2ex] \displaystyle -2v_x=6x^2-6y^2 \\[2ex] \displaystyle v_x=\ -3\ \left(x^2-y^2\right)=-u_y \\[2ex] Subtracting \ those \ equations,\\[2ex] \displaystyle 2v_y=12xy\ \\[2ex] \displaystyle v_y=6xy=\ u_x \\[2ex] \displaystyle f\left(z\right)=u+iv\ ,\ \ \ \\[2ex] \displaystyle Partially\ differentiating\ w.r.t\ x,\\[2ex] f_x^{'}(z)=u_x+iv_x\\[2ex] \displaystyle f_x^{'}=6xy+i\ (-3x^2+3y^2) \\[2ex] Integrating w.r.t x\\ \displaystyle f\left(z\right)=\int6xy+i\ \left(-3x^2+3y^2\right)\ dx\\[2ex] f(z)=\frac{6x^2}{2}y+i\left(-\frac{3x^3}{3}+3xy^2\right)+f\left(y\right) \\[2ex] \displaystyle f\left(z\right)=3x^2y+i\left(3xy^2-x^3\right)+f(y) \ldots{}\ldots{}(D)\\[2ex] \displaystyle f\left(z\right)=u+iv\ ,\ \ \ \\[2ex] $

$\displaystyle Partially\ differentiating\ w.r.t\ y,\\[2ex] f_y^{'}(z)=u_y+iv_y \\[2ex] \displaystyle f_y^{'}=3(x^2-y^2)+i\ (6xy) \\[3ex] \textit{Integrating w.r.t y}\\[2ex] \displaystyle f\left(z\right)=\int3(x^2-y^2)+i\ (6xy)dy \\[2ex] f(z)=3\left(x^2y-\frac{y^3}{3}\right)+i\left(\frac{6xy^2}{2}\right)+f\left(x\right) \\[2ex] \displaystyle f\left(z\right)=3x^2y-y^3+i\left(3xy^2\right)+f\left(x\right)……(E) \\[2ex] $

$\textit{Comparing (D) and (E), } \displaystyle f\left(y\right)=\ -y^3, \\[2ex] \displaystyle f\left(x\right)=\ -ix^3\ \\[2ex] \displaystyle f\left(z\right)=\ 3x^2y-y^3+i\left(3xy^2-x^3\right) \\[2ex] Consider\\[2ex] z^3={\left(x+iy\right)}^3\\[2ex] z^3=x^3+3x^2\left(iy\right)+3x{\left(iy\right)}^2+{\left(iy\right)}^3 \\[2ex] \displaystyle z^3=x^3+3ix^2y-3xy^2-iy^3 \\[2ex] \textit{So, $\[2ex]-iz^3=\ -i\ \left(x^3+3ix^2y-3xy^2-iy^3\right)\[2ex] -iz^3=\ -ix^3+3x^2y+3ixy^2-y^3\[2ex] -iz^3=f\left(z\right)$}\\[2ex] \displaystyle f\left(z\right)=\ -iz^3 \\[2ex]$