if the complex numbers x and y satisfy$x^3-y^3=98i$and  $x-y=7i$then $xy=a+ib$ where $a,b\in R$ the value of $\frac{a+b}{3}=\_\_\_\_\_\_\_$

$x^3-y^3=98i$

$x-y=7i$

Cube on both sides

$\begin{array}{l}\Rightarrow {(x-y)}^3=7i\\ \Rightarrow x^3-y^3-3xy(x-y)=343i^3\\ \Rightarrow 98i-3xy(x-y)=-343i\\ \Rightarrow 98i-21ixy=-343i\\ \Rightarrow 98-21xy=-343\end{array}$

$\begin{array}{l}98+343=21xy\\ \frac{441}{21}=xy\\ xy=21\to \left(1\right)\\ xy=a+ib\to \left(2\right)\\ a=21\\ b=0\\ \frac{a+b}{3}=\frac{21+0}{3}=\frac{21}{3}=7\end{array}$