# It would help in Future

If $x + y + z = a$

$xy + yz + zx = b$

$xyz = c$

then evaluate $(x - y)(y - z)(z - x)$ in terms of a,b,c Note by Dev Sharma
5 years ago

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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.

- 5 years ago

Correct!

- 5 years ago

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 ?

- 5 years ago

we could always use vietas. i remember a problem with a similar question, and a vietas solution. it was, well, too messy...

- 5 years ago

Ok

- 5 years ago

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

by forming a cubic equation

- 5 years ago

@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.

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...

- 5 years ago

yes, its hard to solve cubic equation... And we have to use wolfram alpha

- 5 years ago

or apply cardano method

- 5 years ago

yes.

- 5 years ago

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

- 5 years ago

- 5 years ago

See my comment below.

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

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

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

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

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

- 4 years, 12 months ago

Nice suggestion. Thanks.

- 4 years, 12 months ago