Complex Numbers

$\large \displaystyle x^3 - y^3 = 98i$

$\large \displaystyle \color{#3D99F6} (x-y) \color{#333333} (x^2 + xy + y^2) = 98i$

$\large \displaystyle \color{#3D99F6} 7i \color{#333333} (x^2 + xy + y^2) = 98i$

$\large \displaystyle x^2 + xy + y^2 = 14$

$\large \displaystyle \color{#3D99F6}(x-y)\color{#333333} ^2 + 3xy = 14$

$\large \displaystyle \color{#3D99F6}(7i)\color{#333333} ^2 + 3xy = 14$

$\large \displaystyle 3xy = 63$

$\color{#20A900} \boxed{\large \displaystyle xy = 21 + 0i }$

Then:

$\large \displaystyle \color{#3D99F6} a = 21, b = 0, \boxed{\large \displaystyle \frac{a+b}{3} = 7 }$

$\begin{aligned} x^3-y^3 & = 98i \\ {\color{#3D99F6}(x-y)}(x^2+xy+y^2) & = 98i & \small \color{#3D99F6} \text{Given that }x-y=7i \\ {\color{#3D99F6}7i}(x^2+xy+y^2) & = 98i & \small \color{#3D99F6} \text{Divide both sides by }7i \\ x^2+xy+y^2 & = 14 & ...(1) \end{aligned}$

From:

$\begin{aligned} x-y & = 7i & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 - 2xy + y^2 & = -49 & ...(2) \end{aligned}$

Then, from $(1)-(2): \quad 3xy = 63$ , $\implies xy = 21 + i0$ and $\dfrac {a+b}3 = \dfrac {21+0}3 = \boxed 7$ .