Everybody knows the quadratic formula, which lets you solve this equation: a x2 + b x + c = 0. Turns out it's not hard to solve when there's also an x3 term. There are either one or three solutions. This algorithm makes me happy.
use constant pi => 3.141592653589793; sub cubic { # solve a cubic equation in the form # x^3 + a x^2 + b x + c = 0 my ($a, $b, $c) = @_; my $q = $a*$a/9 - $b/3; my $r = ($a*$a/27 - $b/6)*$a + $c/2; my $s = $a / -3; my $d = $r*$r - $q*$q*$q; if ($d > 0) { my $t = (sqrt($d) + abs($r)) ** (1/3); my $u = ($t + $q / $t); return $r > 0 ? $s - $u : $s + $u; } else { my $t = atan2(sqrt(-$d), $r) / 3; my $u = 2 * sqrt($q); # $d <= 0 implies $q >= 0 return ( $s - $u * cos($t), $s - $u * cos($t + 2/3*pi), $s - $u * cos($t - 2/3*pi), ); } }

In reply to solve cubic equations by no_slogan

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