Question 988414
X^6-x^2=0   
x^2(x^4-1)=0, factoring out an x^2
X^4-1 is a difference of squares.  Looking at that alone.
(x^2+1)(x^2-1)=0
And (x^2-1) is a difference of squares=(x+1)(x-1)
Put it all together:
x^2*(x^2+1)(x+1)(x-1)=0
It has roots of 0 (with multiplicity 2, so there is a "bounce",+/- i, +/-1
{{{graph(300,300,-2,2,-5,5,x^6-x^2)}}}