# 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
4 years, 6 months 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.

- 4 years, 6 months ago

Correct!

- 4 years, 6 months 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 ?

- 4 years, 6 months ago

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

- 4 years, 6 months ago

Ok

- 4 years, 6 months ago

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

- 4 years, 6 months ago

by forming a cubic equation

- 4 years, 6 months 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.

- 4 years, 6 months ago

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...

- 4 years, 5 months ago

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

- 4 years, 5 months ago

or apply cardano method

- 4 years, 5 months ago

- 4 years, 5 months ago

yes.

- 4 years, 5 months 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$=y-\dfrac{coefficeint_{x^2}}{3}$.

- 4 years, 5 months ago

- 4 years, 5 months ago

- 4 years, 5 months ago

See my comment below.

- 4 years, 5 months ago

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, 5 months ago

Nice suggestion. Thanks.

- 4 years, 5 months ago