How do you convert $9 x^{3} - 2 x - 12 y^{2} = 8$ $9x^3-2x-12y^2=8$ into polar form?

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Answer 1 · 2018-03-14T17:54:40

$9 r^{3} {cos}^{3} θ - 2 r cos θ - 12 r^{2} {sin}^{2} θ = 8$ $9r^3cos^3theta-2rcostheta-12r^2sin^2theta=8$

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$9 {(r cos θ)}^{3} - 2 (r cos θ) - 12 {(r sin θ)}^{2} = 8$ $9(rcostheta)^3-2(rcostheta)-12(rsintheta)^2=8$

$9 r^{3} {cos}^{3} θ - 2 r cos θ - 12 r^{2} {sin}^{2} θ = 8$ $9r^3cos^3theta-2rcostheta-12r^2sin^2theta=8$

How do you convert 9x^3-2x-12y^2=89x3−2x−12y2=89x^3-2x-12y^2=8 into polar form?