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

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

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TopNewestLet \(\Delta=(x-y)^2(y-z)^2(z-x)^2\), 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 \((m-n)^2=(m+n)^2-4mn\):

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

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

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

\((x-y)(y-z)(z-x)=\sqrt{a^2b^2-4b^3-4a^3c+18abc-27c^2}\) – Alan Enrique Ontiveros Salazar · 1 year, 9 months ago

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– Dev Sharma · 1 year, 9 months ago

Nice suggestion. Thanks.Log in to reply

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. – Aareyan Manzoor · 1 year, 9 months ago

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– Dev Sharma · 1 year, 9 months ago

Correct!Log in to reply

What If it was unknown to us?

Is there Any other way or it is a considerable one ? – Akshat Sharda · 1 year, 9 months ago

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– Aareyan Manzoor · 1 year, 9 months ago

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

– Akshat Sharda · 1 year, 9 months ago

OkLog in to reply

– Svatejas Shivakumar · 1 year, 9 months ago

Can you find the values for \(x,y,z\)?Log in to reply

– Dev Sharma · 1 year, 9 months ago

by forming a cubic equationLog in to reply

@Dev Sharma Please reply to my comment. – Svatejas Shivakumar · 1 year, 9 months ago

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– Dev Sharma · 1 year, 9 months ago

whats your question?Log in to reply

– Svatejas Shivakumar · 1 year, 9 months ago

See my comment below.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. – Svatejas Shivakumar · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 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...Log in to reply

– Dev Sharma · 1 year, 9 months ago

yes, its hard to solve cubic equation... And we have to use wolfram alphaLog in to reply

– Aareyan Manzoor · 1 year, 9 months ago

or apply cardano methodLog in to reply

– Svatejas Shivakumar · 1 year, 9 months ago

wow I don't know about this cardano method. Can if work for all cubic equations?Log in to reply

– Aareyan Manzoor · 1 year, 9 months ago

yes.Log in to reply

– Aareyan Manzoor · 1 year, 9 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\(=y-\dfrac{coefficeint_{x^2}}{3}\).Log in to reply