How do you convert $3 = {(x - y)}^{2} + x$ $3=(x-y)^2+x$ into polar form?

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Answer 1 · 2016-11-07T23:28:51

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$
$3 = {(r cos θ - r sin θ)}^{2} + r cos θ$ $3=(rcostheta-rsintheta)^2+rcostheta$

How do you convert 3=(x-y)^2+x3=(x−y)2+x3=(x-y)^2+x into polar form?