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
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@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.

@Brilliant Member
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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
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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$=y-\dfrac{coefficeint_{x^2}}{3}$.

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## Comments

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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.

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Correct!

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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 ?

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Can you find the values for $x,y,z$?

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by forming a cubic equation

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@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.

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$(a+b)^3-3ab(a+b)=a^3+b^3$ 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.

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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 $(m-n)^2=(m+n)^2-4mn$:

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

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

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

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

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Nice suggestion. Thanks.

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