Cardano's method: help needed

1. Jan 15, 2007

murshid_islam

i am trying to solve $$x^3 + \frac{5}{3}x - \frac{200}{27} = 0$$ using Cardano's method. i got one solution,

$$x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}$$

now, using a calculator, i can see that x = 5/3. but how can i determine that without using a calculator, i.e., how can i simplify the above expression?

thanks in advance to anybody who can help.

Last edited: Jan 15, 2007
2. Jan 15, 2007

Gib Z

Well since you have one factor, divide the original expression by that to get a quadratic equation, use quadratic forumula to solve for remaining roots.

3. Jan 15, 2007

murshid_islam

that's not what i asked. i know how i can solve for the remaining roots. i wanted to know how i can simplify the following expression:
$$\sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}$$

using a calculator i found that it is equal to 5/3. but how can i find that without using a calculator?

Last edited: Jan 15, 2007
4. Jan 15, 2007

HallsofIvy

Your equation is the same as 27x3+ 45x- 200= 0. Any rational root must have numerator a factor of 200= 2852 and denominator a factor of 27= 33. Try combinations of those until one works. Other than that, there is no "elementary" method to reduce $$x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}$$ to 5/3.

5. Jan 15, 2007

dextercioby

Sure it is

$$x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}=\frac{1}{3}\left(\sqrt[3]{100+45\sqrt{5}} +\sqrt[3]{100-45\sqrt{5}}\right)$$

Denote by "a" the number inside the round bracket. You wish to show that "a=5".

Consider $a^{3}$. By an elementary application of the formula

$(x+y)^{3} \equiv x^{3} +y^{3} +3x^{2}y+3xy^{2}$

one finds out that

$$a^{3}=200-15 a$$

The only viable solution to our initial purpose is a=5.

QED.

Daniel.

6. Jan 15, 2007

murshid_islam

how does one find that out? it involves solving another cubic equation using cardano's method.

7. Jan 15, 2007

dextercioby

No need for Cardano method. You can see very clearly that a=5 is the viable solution by looking at the graph intersection for the curve y=x^3 and y=200-15 x.

They intersect in one point only. x=5 and y=125.

Daniel.

8. Jan 15, 2007

Hurkyl

Staff Emeritus
Apply the rational root theorem: it always gives you all possible rational roots to an integer polynomial. (HoI applied it to your problem)

9. Jan 15, 2007

uart

Well it's easy enough to verify that 5 is the only real solution to that equation which is in effect the same thing as verifying that the original solution is equal to 5/3.

10. Jan 15, 2007

murshid_islam

thanks a lot. i can of course use the rational root theorem. i just wanted to know if i used Cardano's method, whether there is a way to simplify
$$\sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}$$ to $$\frac{5}{3}$$ directly.

anyway, thanks a lot guys.

Last edited: Jan 15, 2007
11. Jan 15, 2007

HallsofIvy

Very nice. Thank you

12. Jan 16, 2007

robert Ihnot

Y=$$\sqrt[3]{2+\sqrt5} +\sqrt[3]{2-\sqrt5 }$$ Niederhoffer, writing in "The Education of a Speculator," got this problem on a high school math honor course. He was to chose a simple answer for this. Among the choices, he just guessed 1, which was correct.

The dextercioby method works well on that. Just set it equal to Y and cube. Then replace the middle part by -Y. The result is Y^3 = 4-3Y, making 1 an easy choice.

Last edited: Jan 16, 2007