# Cubic roots and Cardano formula

On solving the cubic equations, applying Cardano formula yield complex results. I wanted to evaluate the exact roots (not numerical) but I ended up with complex numbers/nested radicals. To get rid of complex numbers/nested radicals, we are either to denest the nested radical or use trigonometric formula for cubic equations. The trigonometric formula has to be avoided because it won’t give exact roots. So we are left to denest the nested radical of the form

1. $\sqrt{a+b\sqrt{c}}$
2. $\sqrt{a+b\sqrt{c}} +\sqrt{a-b\sqrt{c}}$

In the first case I was asked to “assume” the solution which is really annoying, for the second I was told to suppose the radical to be equal to “$x$” and then cubing on both sides gives the same cubic equation, now “assume” the solution. I seem to be processing in a cycle, I am greatly troubled.

• In what way are complex numbers and nested radicals not exact? – Zev Chonoles Aug 14 '13 at 6:31
• The trig formulas do give exact roots, just not in the form you prefer. – André Nicolas Aug 14 '13 at 6:35
• What you most likely mean is that you want exact symbolic expressions for the real and imaginary parts independently, rather than one whole expression for the complex number that may obscure what the real/imaginary parts are. Can you elaborate on the "assume the solution" method you've been asked to do, perhaps with an example? – anon Aug 14 '13 at 6:44
• – mercio Aug 14 '13 at 6:48
• 1) Not exact signifies "numerical approximations" 2) The trig formulae give "numerical approximations" and not in "irrational form" of roots sought. 3) Consider the equation x^3+3x=4 X=∛(2+√5) +∛(2-√5) Cubing both sides: x^3+3x-4=0 “assuming” x=1 anf equating the sides will prove that x=1 ): This is the so called “assuming method” I showed – Hashir Omer Aug 14 '13 at 7:58