This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
let x,y,z be roots of P(k)=k3−ak2+bk−c, then (x−y)(y−z)(z−x)=(x−y)2(y−z)2(z−x)2=Discriminant
we know the discriminant formulas, insert it and:
(x−y)(y−z)(z−x)=−4a3c+a2b2+18abc−4b3−27c2 solved.
@Dev Sharma
–
@Dev Sharma Is it using Vieta's formula? And after that how would you solve the cubic equation? It looks hard to solve the simultaneous equation since the first equation is of degree 1, the second equation is of degree 2 and the third equation is of degree 3.
@A Former Brilliant Member
–
If you want to solve a cubic equation , first look for rational roots by rational root theorem. If a rational root exists , with the help of factor theorem , obtain a factor and use synthetic division to obtain other factor(s). If rational roots don't exist , you need to use Cardano's method...
@Aareyan Manzoor
–
cardano basically is like this:
(a+b)3−3ab(a+b)=a3+b3 let a+b=x
x3−3abx−a3−b3=0
so in the cubic x3+px+q=0,x=a+b,p=−3ab,q=−a3−b3 this will form a tri-quadratic in terms of p and q which we can solve to get our solutions. you can turn any cubic like this by putting x=y−3coefficeintx2.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
let x,y,z be roots of P(k)=k3−ak2+bk−c, then (x−y)(y−z)(z−x)=(x−y)2(y−z)2(z−x)2=Discriminant we know the discriminant formulas, insert it and: (x−y)(y−z)(z−x)=−4a3c+a2b2+18abc−4b3−27c2 solved.
Log in to reply
Correct!
Log in to reply
Here, it requires a pre-requisite knowledge about the discriminant.
What If it was unknown to us?
Is there Any other way or it is a considerable one ?
Log in to reply
we could always use vietas. i remember a problem with a similar question, and a vietas solution. it was, well, too messy...
Log in to reply
Ok
Can you find the values for x,y,z?
Log in to reply
by forming a cubic equation
Log in to reply
@Dev Sharma Is it using Vieta's formula? And after that how would you solve the cubic equation? It looks hard to solve the simultaneous equation since the first equation is of degree 1, the second equation is of degree 2 and the third equation is of degree 3.
Log in to reply
If you want to solve a cubic equation , first look for rational roots by rational root theorem. If a rational root exists , with the help of factor theorem , obtain a factor and use synthetic division to obtain other factor(s). If rational roots don't exist , you need to use Cardano's method...
yes, its hard to solve cubic equation... And we have to use wolfram alpha
Log in to reply
or apply cardano method
Log in to reply
wow I don't know about this cardano method. Can if work for all cubic equations?
Log in to reply
yes.
Log in to reply
cardano basically is like this: (a+b)3−3ab(a+b)=a3+b3 let a+b=x x3−3abx−a3−b3=0 so in the cubic x3+px+q=0,x=a+b,p=−3ab,q=−a3−b3 this will form a tri-quadratic in terms of p and q which we can solve to get our solutions. you can turn any cubic like this by putting x=y−3coefficeintx2.
@Dev Sharma Please reply to my comment.
Log in to reply
whats your question?
Log in to reply
See my comment below.
Let Δ=(x−y)2(y−z)2(z−x)2, then expand it:
Δ=(xy2+yz2+zx2−(x2y+y2z+z2x))2
Now let m=xy2+yz2+zx2 and n=x2y+y2z+z2x. So, Δ=(m−n)2.
Now, here comes the trick, first find m+n and mn:
m+n=xy(x+y)+xz(x+z)+yz(y+z)=(x+y+z)(xy+xz+yz)−3xyz=ab−3c
mn=(xy2+yz2+zx2)(x2y+y2z+z2x)=(xy)3+(xz)3+(yz)3+3(xyz)2+xyz(x3+y3+z3)
Use the useful identity x3+y3+z3=(x+y+z)3−3(x+y+z)(xy+xz+yz)+3xyz two times to simplify that:
mn=(xy+xz+yz)3−3(xy+xz+yz)(xyz)(x+y+z)+3(xyz)2+3(xyz)2+xyz((x+y+z)3−3(x+y+z)(xy+xz+yz)+3xyz)
mn=b3−3abc+6c2+c(a3−3ab+3c)
mn=b3−6abc+9c2+a3c
Finally, use the fact that (m−n)2=(m+n)2−4mn:
Δ=(ab−3c)2−4(b3−6abc+9c2+a3c)
Δ=a2b2−6abc+9c2−4b3+24abc−36c2−4a3c
Δ=a2b2−4b3−4a3c+18abc−27c2
(x−y)(y−z)(z−x)=a2b2−4b3−4a3c+18abc−27c2
Log in to reply
Nice suggestion. Thanks.