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

let x,y,z be roots of $P(k)=k^3-ak^2+bk-c$, then $(x-y)(y-z)(z-x)=\sqrt{(x-y)^2(y-z)^2(z-x)^2}=\sqrt{Discriminant}$
we know the discriminant formulas, insert it and:
$(x-y)(y-z)(z-x) = \sqrt{-4 a^3 c+a^2 b^2+18 a b c-4 b^3-27 c^2}$ 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:
$(a+b)^3-3ab(a+b)=a^3+b^3$ let a+b=x
$x^3-3abx-a^3-b^3=0$
so in the cubic $x^3+px+q=0,x=a+b,p=-3ab,q=-a^3-b^3$ 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$=y-\dfrac{coefficeint_{x^2}}{3}$.

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_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestlet x,y,z be roots of $P(k)=k^3-ak^2+bk-c$, then $(x-y)(y-z)(z-x)=\sqrt{(x-y)^2(y-z)^2(z-x)^2}=\sqrt{Discriminant}$ we know the discriminant formulas, insert it and: $(x-y)(y-z)(z-x) = \sqrt{-4 a^3 c+a^2 b^2+18 a b c-4 b^3-27 c^2}$ 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

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

$(a+b)^3-3ab(a+b)=a^3+b^3$ let a+b=x $x^3-3abx-a^3-b^3=0$ so in the cubic $x^3+px+q=0,x=a+b,p=-3ab,q=-a^3-b^3$ 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$=y-\dfrac{coefficeint_{x^2}}{3}$.

cardano basically is like this:Log in to reply

@Dev Sharma Please reply to my comment.

Log in to reply

Log in to reply

Log in to reply

Let $\Delta=(x-y)^2(y-z)^2(z-x)^2$, then expand it:

$\Delta=(xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x))^2$

Now let $m=xy^2+yz^2+zx^2$ and $n=x^2y+y^2z+z^2x$. So, $\Delta=(m-n)^2$.

Now, here comes the trick, first find $m+n$ and $mn$:

$m+n=xy(x+y)+xz(x+z)+yz(y+z)=(x+y+z)(xy+xz+yz)-3xyz=ab-3c$

$mn=(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)=(xy)^3+(xz)^3+(yz)^3+3(xyz)^2+xyz(x^3+y^3+z^3)$

Use the useful identity $x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$ two times to simplify that:

$mn=(xy+xz+yz)^3-3(xy+xz+yz)(xyz)(x+y+z)+3(xyz)^2+3(xyz)^2+xyz((x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz)$

$mn=b^3-3abc+6c^2+c(a^3-3ab+3c)$

$mn=b^3-6abc+9c^2+a^3c$

Finally, use the fact that $(m-n)^2=(m+n)^2-4mn$:

$\Delta=(ab-3c)^2-4(b^3-6abc+9c^2+a^3c)$

$\Delta=a^2b^2-6abc+9c^2-4b^3+24abc-36c^2-4a^3c$

$\Delta=a^2b^2-4b^3-4a^3c+18abc-27c^2$

$(x-y)(y-z)(z-x)=\sqrt{a^2b^2-4b^3-4a^3c+18abc-27c^2}$

Log in to reply

Nice suggestion. Thanks.

Log in to reply