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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Cubic equation coefficients from 4 points
For a cubic curve (Bezier) of the form: $ax^3 + bx^2 + cx + d = y$.
I have a given set of four points $P_0, P_1, P_2, P_3$. Such that, $P_0$ is the origin and the other three are equidistant along the ...
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How to draw graph of cubic function
I am taking a course in calculus and wanted to refresh my memory before the semester starts. And I have been working on drawing graphs from cubic functions. I am not that experienced with LaTex and ...
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Minimizing a cubic polynomial over $\Bbb N$ [closed]
Let the polynomial function $f : \Bbb N \to \Bbb N$ be defined by
$$f (M) := 2M^3N + 2M^3 - M^2N^2 - 3M^2N + 2M^2 - MN^3 + MN^2 - 2MN + \frac{N^4}{2} + \frac{N^2}{2}$$
where $N$ is given natural ...
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The Diophantine equation $P_1^3 + P_2^3 + P_3^3 = P_4^3$
Consider the Diophantine equation
$$P_1^3 + P_2^3 + P_3^3 = P_4^3$$
Where $P_n$ are distinct odd primes.
What are the smallest solutions ?
Is there even a solution ?
Or is there a reason no such ...
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Failing to solve cubic equation
I'm trying to solve a more complex cubic equation but to simplify things as a start I picked this one:
$$ 3\cdot 4^3+2\cdot 4-200=0 $$
Here $x$ is $4$.
I'm looking at wikipedia and trying to solve ...
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cubic equation edge cases
Working on general cubic equation solver in form ax^3+bx^2+cx+d=0 And have no clue for special cases:
In terms of cubic there should be one real root and two complex, or 3 real roots if coefficients ...
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Prove that $a_3 \lambda^{3} + a_2 \lambda^{2} + a_1 \lambda + a_0 = 0$ has three real roots
I'm trying to prove that the cubic equation
$a_3 \lambda^{3} + a_2 \lambda^{2} + a_1 \lambda + a_0 = 0$
has three real roots. The coefficients are
$a_3 = - 1 - \sigma - \tau - \chi$
$a_2 = -2 (\sigma +...
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There is a compass-like tool that can draw $y=x^2$ on paper. Is there one for $y=x^3$?
Is there a tool that can draw $y=x^3$ on paper?
I'm referring to low-tech tools, e.g. not computers.
I only know of tools that can draw $y=x^2$. The YouTube video "Conic Sections Compass" ...
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Prove this potential cubic theorem/formula [closed]
Prove that $\dfrac{x³+y³+z³}{x+y+z}=x²+y+z$; if $x<y<z$; $y=x+1$; $z=y+1$; and $x$, $y$ and $z$ are positive whole numbers.
If you prove this, I technically discovered a new formula since I ...
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Prove $2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a)$ for $abc=1.$
Let $a,b,c>0: abc=1.$ Prove that: $$2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a). $$
I've tried to use a well-known lemma but the rest is quite complicated for me.
...
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Three real roots of a cubic
Question: If the equation $z^3-mz^2+lz-k=0$ has three real roots, then necessary condition must be _______
$l=1$
$ l \neq 1$
$ m = 1$
$ m \neq 1$
I know there is a question here on stack about ...
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Why are all real inflection points on a cubic projective algebraic curve on 1 line?
Say we have $C\subset \mathbb{CP}^2$, a smooth curve of degree 3. I am aware of the group structure on cubics, what I don't get, is why are all inflection points with only real coordinates lie on a ...
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More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$
I. Solutions
In a previous post, On sums of three cubes of form $a^3+b^3+c^3=(c+1)^3$, an example of which is the well-known,
$$3^3+4^3+5^3=6^3$$
we asked if there were polynomial parameterizations ...
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Disjoint exceptional lines on non-minimal cubic surface
A diagonal cubic surface $\sum_{i=0}^3a_iT_i^3=0$ is not minimal if, for example, $a_1a_2a_3^{-1}a_4^{-1}\in(k^*)^3$. This should be because there is an exceptional line $D$ such that no element in ...
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Prove $(1+a^3) (1+b^3)(1+c^3) \ge (\frac{ab+bc+ca+1}{2})^3$
This is a question from 4U maths (highest level of Y12 maths in Australia) from a generally difficult paper. The question itself does not define what a, b, and c are - based on the comments, we assume ...
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