What is represented by the equation ${x^3} + {y^3} + \left( {x + y} \right)\left( {xy - ax - ay} \right) = 0 $?

Hint- Here, we will be using the formula ${x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} + {y^2} - xy} \right)$ in the given equation and then we will compare the equations obtained with the general equation of straight line $y = mx + c$ and that of circle ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}$.

Complete step-by-step solution -
The given equation is ${x^3} + {y^3} + \left( {x + y} \right)\left( {xy - ax - ay} \right) = 0$
Using the formula ${x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} + {y^2} - xy} \right)$ , we get
$
   \Rightarrow \left( {x + y} \right)\left( {{x^2} + {y^2} - xy} \right) + \left( {x + y} \right)\left( {xy - ax - ay} \right) = 0 \\
   \Rightarrow \left( {x + y} \right)\left[ {\left( {{x^2} + {y^2} - xy} \right) + \left( {xy - ax - ay} \right)} \right] = 0 \\
   \Rightarrow \left( {x + y} \right)\left[ {{x^2} + {y^2} - ax - ay} \right] = 0 \\
 $
From above equation we can say that either of the two terms on the LHS is equal to 0
i.e., either $\left( {x + y} \right) = 0$ or $\left[ {{x^2} + {y^2} - ax - ay} \right] = 0$
$ \Rightarrow y = - x{\text{ }} \to {\text{(1)}}$ or
\[
  {x^2} - ax + {\left( {\dfrac{a}{2}} \right)^2} + {y^2} - ay + {\left( {\dfrac{a}{2}} \right)^2} - {\left( {\dfrac{a}{2}} \right)^2} - {\left( {\dfrac{a}{2}} \right)^2} = 0 \\
   \Rightarrow {\left( {x - \dfrac{a}{2}} \right)^2} + {\left( {y - \dfrac{a}{2}} \right)^2} - 2{\left( {\dfrac{a}{2}} \right)^2} = 0 \\
   \Rightarrow {\left( {x - \dfrac{a}{2}} \right)^2} + {\left( {y - \dfrac{a}{2}} \right)^2} = 2{\left( {\dfrac{a}{2}} \right)^2} \\
   \Rightarrow {\left( {x - \dfrac{a}{2}} \right)^2} + {\left( {y - \dfrac{a}{2}} \right)^2} = {\left( {\dfrac{{a\sqrt 2 }}{2}} \right)^2} \\
   \Rightarrow {\left( {x - \dfrac{a}{2}} \right)^2} + {\left( {y - \dfrac{a}{2}} \right)^2} = {\left( {\dfrac{a}{{\sqrt 2 }}} \right)^2}{\text{ }} \to {\text{(2)}} \\
 \]
As we know that general equation of any straight line is $y = mx + c{\text{ }} \to {\text{(3)}}$, where m is the slope of the straight line and c is the y-intercept of the straight line.
Also, we know that the equation of any circle with centre coordinate as $\left( {{x_1},{y_1}} \right)$ and radius of the circle as r is given by ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}{\text{ }} \to {\text{(4)}}$
Now on comparing equations (1) and (3), we can say that $y = - x$ is the equation of a straight line passing through origin with a slope of -1.
Also on comparing equations (2) and (4), we can say that \[{\left( {x - \dfrac{a}{2}} \right)^2} + {\left( {y - \dfrac{a}{2}} \right)^2} = {\left( {\dfrac{a}{{\sqrt 2 }}} \right)^2}\] is the equation of a circle with centre coordinate as $\left( {\dfrac{a}{2},\dfrac{a}{2}} \right)$ and radius of $\dfrac{a}{{\sqrt 2 }}$.
Therefore, the given equation ${x^3} + {y^3} + \left( {x + y} \right)\left( {xy - ax - ay} \right) = 0$ represents a circle and a straight line.

Note- In these types of problems, we simplify the given equation in such a way that the simplified equation or equations refer to a general form of equation of some known line or curve. Here, the given combined equation resembles the general equation of a line and a circle.