help calc!!!!!! plz?

Question

test for symmetry(x-axis,y-axis,origin)
5x^6 + 3xy^2-y^6=7

Answer 1

This is symmetric around the x axis because the only "y" terms could be rewritten as terms with y^2, and since y^2 is the same whether y is positive or negative, this is symmetric around the x axis.
It is NOT symmetric around the y axis. To prove this all we need to do is pick a point. Let y = 1, and then 5x^6 + 3x = 8, and x=1 is a solution. But, x = -1 is NOT a solution.

If the function is to be symmetric around the origin, then since (1,1) is a solution, (-1,-1) also must be a solution, but it is not.

Answer 2

/1/ So 5x^6 + 3xy^2 -y^6 –7=0; hence x= (y^6 -5x^6 +7)/(3y^2), thus y(x) is symmetrical with respect to x-axis since all functions in RHS are even!
\2\ Now y^2 = (7 -5x^6) / (3x –y^4) is not = (7-5x^6) / (-3x –y^4), thus y(x) is not symmetrical wrt y-axis!
/3/ y(x) is not symmetrical wrt origin because of /1/ and \2\;