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

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

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

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

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

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Nice suggestion. Thanks.

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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.

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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 Please reply to my comment.

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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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