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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
1 year, 5 months ago

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

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

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

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

\((x-y)(y-z)(z-x)=\sqrt{a^2b^2-4b^3-4a^3c+18abc-27c^2}\) Alan Enrique Ontiveros Salazar · 1 year, 5 months ago

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@Alan Enrique Ontiveros Salazar Nice suggestion. Thanks. Dev Sharma · 1 year, 5 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. Aareyan Manzoor · 1 year, 5 months ago

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@Aareyan Manzoor Correct! Dev Sharma · 1 year, 5 months ago

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@Dev Sharma 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 ? Akshat Sharda · 1 year, 5 months ago

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@Akshat Sharda we could always use vietas. i remember a problem with a similar question, and a vietas solution. it was, well, too messy... Aareyan Manzoor · 1 year, 5 months ago

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@Aareyan Manzoor Ok Akshat Sharda · 1 year, 5 months ago

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@Aareyan Manzoor Can you find the values for \(x,y,z\)? Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar by forming a cubic equation Dev Sharma · 1 year, 5 months ago

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@Dev Sharma @Dev Sharma Please reply to my comment. Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar whats your question? Dev Sharma · 1 year, 5 months ago

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@Dev Sharma See my comment below. Svatejas Shivakumar · 1 year, 5 months ago

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@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. Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar 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... Nihar Mahajan · 1 year, 5 months ago

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@Svatejas Shivakumar yes, its hard to solve cubic equation... And we have to use wolfram alpha Dev Sharma · 1 year, 5 months ago

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@Dev Sharma or apply cardano method Aareyan Manzoor · 1 year, 5 months ago

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@Aareyan Manzoor wow I don't know about this cardano method. Can if work for all cubic equations? Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar yes. Aareyan Manzoor · 1 year, 5 months ago

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@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\(=y-\dfrac{coefficeint_{x^2}}{3}\). Aareyan Manzoor · 1 year, 5 months ago

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