Are there any symmetries in the graph #2y^3=7x^3+5#?

1 Answer
George C.
Feb 8, 2017

As a real valued function, no.
As a complex valued relation, yes.

Explanation:

Given:

#2y^3 = 7x^3+5#

There are no symmetries as a Real valued function of Real numbers.

First note that this does define a function:

#y = root(3)(7/2x^3+5/2)#

Secondly, this is neither odd nor even. For example:

#root(3)(7/2color(red)((1))^3+5/2) = root(3)(6)#

#root(3)(7/2color(red)((-1))^3+5/2) = root(3)(-1)#

Thirdly, it is asymptotic to the line:

#y = root(3)(7/2)x#

graph{2y^3 = 7x^3+5 [-10, 10, -5, 5]}

#color(white)()#
Footnote

There are two axes of order three rotational symmetry as a Complex valued relation of Complex numbers, but it is quite hard to visualise the four dimensional geometry involved.

For each value of #(x, y)# satisfying:

#2y^3 = 7x^3+5#

all of the following #9# points satisfy the relation:

#(x, y)#, #(omega x, y)#, #(omega^2 x, y)#

#(x, omega y)#, #(omega x, omega y)#, #(omega^2x, omega y)#

#(x, omega^2 y)#, #(omega x, omega^2 y)#, #(omega^2x, omega^2 y)#

where #omega = -1/2+sqrt(3)/2i = cos ((2pi)/3) + i sin((2pi)/3)# is the primitive complex cube root of #1#.

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