What are represented by the equation ${{x}^{3}}-{{y}^{3}}=\left( y-a \right)\left( {{x}^{2}}-{{y}^{2}} \right)$.

Hint: Substitute the basic equations in place of $\left( {{x}^{3}}-{{y}^{3}} \right)$and$\left( {{x}^{2}}-{{y}^{2}} \right)$.
Simplify them and equate them to zero. You will get the expression for a straight line and parabola.

Complete step-by-step answer:
We are given the equation
${{x}^{3}}-{{y}^{3}}=\left( y-a \right)\left( {{x}^{2}}-{{y}^{2}} \right)\ldots \ldots (1).$
We know the basic expressions
$\begin{align}
  & {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right) \\
 & {{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right) \\
\end{align}$
Substitute these in equation (1).
\[\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)=\left( y-a \right)\left( x-y \right)\left( x+y \right)\]
From this, \[x-y=0\]
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}+xy=\left( y-a \right)\left( x-y \right)\ldots \ldots (2). \\
\end{align}\]
Here we have expanded the value of $( y-a) (x-y)$.
 $\Rightarrow\left( y-a \right)\left( x-y \right)=xy-{{y}^{2}}-ax+ay \\ $
Thus substitute the value of \[\left( y-a \right)\left( x+y \right)\] in equation (2).
\[\Rightarrow {{x}^{2}}+{{y}^{2}}+xy=xy+{{y}^{2}}-ax-ay\]
Now cancel \[xy\] and \[{{y}^{2}}\]from LHS and RHS of the above expression and rearrange the equations.
\[\begin{align}
  & {{x}^{2}}=-ay-ax \\
 & \Rightarrow {{x}^{2}}+ax+ay=0 \\
\end{align}\]
Thus we obtained 2 curves, \[x-y=0\] and \[{{x}^{2}}+ax+ay=0\]
\[x-y=0\], represents a straight line.
\[{{x}^{2}}+ax+ay=0\], represents a parabola.
Thus, the given equation represents a straight line and parabola.

Note: You might cancel out \[\left( x-y \right)\] in the initial stages of the problem. Remember that \[x-y=0\]represents the equation of a straight line. Remember the basic formulas that we have used here. Without the basic equations you can't simplify the given expression. The equation \[{{x}^{2}}+ax+ay=0\] is said to be a quadratic equation as the degree of equation is 2. When we draw a graph for this equation, we get a parabola. The graph of a quadratic equation in two variables (y = ax² + bx + c ) is called a parabola and the equation \[x-y=0\] is said to be a linear equation in one variable as the degree of equation is 2. When we draw a graph for this equation, we get a straight line.