The equation $${\left( {{x^2} - {a^2}} \right)^2}{\left( {{x^2} - {b^2}} \right)^2} + {c^4}{\left( {{y^2} - {a^2}} \right)^2} = 0$$         represents $$\left( {c \ne 0} \right)$$

Solution :
$${\left( {{x^2} - {a^2}} \right)^2}{\left( {{x^2} - {b^2}} \right)^2} + {c^4}{\left( {{y^2} - {a^2}} \right)^2} = 0$$
This being the sum of two perfect squares, each term must be zero.
Hence, we get
$$\eqalign{ & {\left( {{x^2} - {a^2}} \right)^2}{\left( {{x^2} - {b^2}} \right)^2} = 0 \cr & {\text{or }}\left( {{x^2} - {a^2}} \right)\left( {{x^2} - {b^2}} \right) = 0 \cr & {\text{or }}\left( {x - a} \right)\left( {x + a} \right)\left( {x - b} \right)\left( {x + b} \right) = 0......\left( 1 \right) \cr & {\text{and }}{c^4}{\left( {{y^2} - {a^2}} \right)^2} = 0 \cr & {\text{or }}{c^2}\left( {{y^2} - {a^2}} \right) = 0 \cr & {\text{or }}{c^2}\left( {y + a} \right)\left( {y - a} \right) = 0......\left( 2 \right) \cr} $$
Equation number $$\left( 1 \right)$$ holds good for $$x = \pm a{\text{ or }}x = \pm b$$
Equation number $$\left( 2 \right)$$ is satisfied by $$y = \pm a$$
As both of these should be simultaneously satisfied, the given equation represents $$8$$ points which we get as a result of different combinations of $$\left( 1 \right)$$ and $$\left( 2 \right),$$ namely $$\left( { \pm a,\, \pm a} \right),\,\left( { \pm b,\, \pm a} \right).$$