Solve x^3 + 5/3x - 200/27 = 0 with Cardano's Method: Tips & Tricks

[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.

 

[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.

 

[murshid_islam]

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.

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?

 

[HallsofIvy]

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.

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.

 

[dextercioby]

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.

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.

 

[murshid_islam]

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.

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

 

[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.

 

[Hurkyl]

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

 

[uart]

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

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.

 

[murshid_islam]

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

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.

 

[HallsofIvy]

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.

Very nice. Thank you

 

[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.