# How am I supposed to separate the variables in this polar converted equation?

I have attacked this equation from many sides and I can't figure out how to get 'r' by itself in the rectangular to polar conversion.

$$\ x^3+xy+y^2=5$$ What makes this problem difficult is the 'xy' and 'x^3' terms making it difficult to simplify once the x's and y's are substituted with sin's and cosines'

I have a feeling there is a substitution that I could do somewhere because otherwise, I can't separate the r's and theta's

This is the closest I've gotten:

subtracting x^3 from both sides: $$\ xy+y^2=5-x^3$$

convert: $$\ r^2cos(\theta)sin(\theta)+r^2sin^2(\theta)=5-r^3cos^3(\theta)$$

divide both sides by r^2: $$\ cos(\theta)sin(\theta)+sin^2(\theta)= (5/r^2)-rcos^3(\theta)$$

• My guess it's a typo, and x should be squared not cubed Commented May 12, 2019 at 1:40
• So your saying I can't solve for r here? Commented May 12, 2019 at 1:42
• It's highly unlikely that you have $x^3+xy+y^2$ instead of $x^2+xy+y^2$ because of the identity $(x-y)(x^2+xy+y^2) = x^3 - y^3$. Would you mind posting the original question and all the steps you have completed? Commented May 12, 2019 at 1:46
• You don't have enough information. You need something else to tie $x$ and $y$. For example, assuming that your original equation is correct, I can get $x=0$ and $y=\sqrt 5$ or $x=\sqrt[3]5$ and $y=0$ or $y=-2$ and $x=-1$ Commented May 12, 2019 at 1:46
• I have added my steps, the question just wants me to find the equation in polar coordinates. Commented May 12, 2019 at 1:58

As @Andrei commented, I suppose a typo and that $$x^3$$could be $$x^2$$ instead.
Otherwise, solving the cubic equation in $$r$$, the discriminant would be $$\Delta=20 \sin ^3(\theta )\left(\sin (\theta )+ \cos (\theta )\right)^3-675 \cos ^6(\theta )$$ which, for the range $$0 \leq \theta \leq 2\pi$$, would be positive (then three real roots in $$r$$) for $$0.938845 < \theta < 1.97069$$