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Can you solve for y in terms of x?

z=0 by the way

x = cos^3(t) y= sin^3(t) z = 0 right?

its a cubic equation that crosses the x and y axis 3 times.... i see that much

r^3 (cos^3 + sin^3) = sin^2 + cos^2

yeah

I see where the ^1/3 is coming from.... just how to get it :)

it seems as though thats not too far away

yeah, the solution being correct is the easy part ;)

1 is also the "radius" or vector measure from the origin.... we could try that route.

are there any trig identitied we could utilize for this?

ermm, not that i can think of

r^3 (cos^3 + sin^3) = r^2 ; r^2 = 1 so r^3 = 1 ?

ye, i think r^3 must have to be 1

cos^3 + sin^3 = cos^2 + sin^2

cos^3 - cos^2 = sin^3 - sin^3 ??

typoed it...but you can see that lol

ye, i see lol...what do you have to do to a square power to make it ^2/3?
do you cube root it?

yes; ^2/3 means cbrt(^2)

ok, i think i get it, thanks so much for the help!