Questions tagged [cubics]
This tag is for questions relating to cubic equations, these are polynomials with $~3^{rd}~$ power terms as the highest order terms.
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Expressing the roots of a cubic as polynomials in one root
All roots of $8x^3-6x+1$ are real. (*)
The discriminant of $8x^3-6x+1$ is $5184=72^2$ and so the splitting field of $8x^3-6x+1$ has degree $3$.
Therefore, all three roots can be expressed as ...
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Is there anything like “cubic formula”?
Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.
Similarly if we have a cubic ...
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Question Regarding Cardano's Formula
In Cardano's derivation of a root of the cubic polynomial $f(X)=X^3+bX+c$ he splits the variable $X$ into two variables $u$ and $v$ together with the relationship that $u+v=X$. From this he finds that ...
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Is there really analytic solution to cubic equation?
I know the formulas for solving cubic equation, but when I try to use them in both Cardano's method and https://en.wikipedia.org/wiki/Cubic_function#Algebraic_solution, I usually encounter something ...
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On the trigonometric roots of a cubic
We have,
$$x^3+x^2-2x-1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, (6^k)}{7}\Bigr)\\
x^3+x^2-4x+1=0,\quad\quad x =\sum_{k=1}^{4}\,\exp\Bigl(\tfrac{2\pi\, i\, (5^k)}{13}\Bigr)\\
x^3+x^...
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11
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Where is my error in trying to find Pythagorean triples with matching areas?
I'm trying to craft a formula for finding matching areas in Pythagorean triples the way I have done with matching sides and matching perimeters and matching area:perimeter ratios. For example:
$f(10,...
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The sum of three integer cubes $x^3+y^3+z^3 \not\equiv \pm4 \pmod{\!9}$
I'm reading an article about numbers that can be expressed as the sum of three cubes:
It states in the article that if you take the following expressions for $x^3+y^3+z^3 = k$ for $x,y,z,k \in \mathbb{...
2
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2
answers
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How to find the complex roots of $y^3-\frac{1}{3}y+\frac{25}{27}$
I've been trying to solve this for hours and all found was the real solution by Cardano"s formula.
I vaguely remember that if $\alpha$ is a root of a complex number, the other roots are $\omega \...
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27
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1
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Using Vieta's theorem for cubic equations to derive the cubic discriminant
Background:
Vieta's Theorem for cubic equations says that if a cubic equation $x^3 + px^2 + qx + r = 0$ has three different roots $x_1, x_2, x_3$, then
$$\begin{eqnarray*}
-p &=& x_1 + x_2 +...
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5
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Solving $\cos(3x) = \cos(2x)$
I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x)...
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8
votes
2
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Solving a cubic polynomial equation.
Overview
I have tried finding a solution to this problem myself and I have failed. It's just a challenge for me. Could you please tell me how far am I in solving this question?
My approach for ...
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Motivation of Vieta's transformation
The depressed cubic equation $y^3 +py + q = 0$ can be solved with Vieta's transformation (or Vieta's substitution)
$y = z - \frac{p}{3 \cdot z}.$
This reduces the cubic equation to a quadratic ...
- 165k
9
votes
3
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Is there a systematic way of solving cubic equations?
According to my text book, to solve cubic equations, I need to
By trial & error find what value $a$ will make the cubic $0$. The factor will be $(x-a)$.
Then the other factor will be $Ax^2+Bx+C$ ...
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3
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2
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Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
Problem :
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\...
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2
votes
2
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Deriving expression for general cubic equation solution.
This is for the page #1,page #2, page #3, page #4 of the book by Dickson, titled "Introduction to Theory of Algebraic Equations", in which there is derivation of solution for general cubic equation.
...
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