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I would like to be able to simplify the following expression, if it is at all possible.

$$-\left(\frac{A}{3a} + \frac{B}{3b} + \frac{C}{3c}\right) \pm \frac{\sqrt{A^{2}(b^{2}c^{2}) + B^{2}(a^{2}c^{2}) + C^{2}(a^{2}b^{2}) - AB(abc^{2}) - AC(ab^{2}c) - BC(a^{2}bc)}}{3abc},$$

Thank you.

Edit

I believe that I can simplify it to the following, but I am not sure if this is the furthest that one can go to simplify the expression.

$$-\frac{1}{3}\left(\left(\frac{A}{a} + \frac{B}{b} + \frac{C}{c}\right) \mp \sqrt{\left(\frac{A^{2}}{a^{2}} + \frac{B^{2}}{b^{2}} + \frac{C^{2}}{c^{2}} - \frac{AB}{ab} - \frac{AC}{ac} - \frac{BC}{bc}\right)}\right),$$

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  • $\begingroup$ That's probably as nice as it will get. You can simplify the denominator in the 6 terms in the root, maybe that will look nicer. $\endgroup$
    – Jeb
    Jun 15, 2015 at 15:14
  • $\begingroup$ @Jeb How do I do that? Would you be able to write an answer to show me? $\endgroup$
    – Taylor
    Jun 15, 2015 at 15:15
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    $\begingroup$ Use reduced variables $\alpha=A/a\cdots$: $$\frac13\left(-\left({\alpha} + {\beta} + {\gamma}\right) \pm\sqrt{\alpha^{2} + \beta^{2} + \gamma^{2} + \alpha\beta + \alpha\gamma + \beta\gamma}\right)$$ $\endgroup$
    – user65203
    Jun 15, 2015 at 15:19
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    $\begingroup$ . . . It now appears that a correct version of what I had in mind goes like this: $2x^2y^2 + 2x^2z^2 + 2y^2z^2 - x^4 - y^4 - z^4$ $=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)\vphantom{\dfrac11}$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Jun 15, 2015 at 17:25
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    $\begingroup$ Yes. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Jun 15, 2015 at 17:26

1 Answer 1

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HINT:You can go to the form $3X= -(\alpha + \beta+\gamma) \pm\sqrt{(\alpha + \beta + \gamma)^2 – (\alpha\beta + \alpha\gamma + \beta\gamma)}$ and this is actually six times the roots of

$X^2 +(\alpha + \beta+\gamma) X + \frac{(\alpha\beta + \alpha\gamma + \beta\gamma)}{4}=0$.

I fear you could not simplify without additional conditions on your letters $A,B,C,a,b,c$.

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  • $\begingroup$ Thank you very much for your answer, but I am afraid I do not understand how you have managed to convert the expression to a quadratic. $\endgroup$
    – Taylor
    Jun 15, 2015 at 16:53
  • $\begingroup$ Try (with the formula you well know) to solve the quadratic equation I gave you and see at the transformation gave by Yves Daoust which is more or less the same that mine $\endgroup$
    – Piquito
    Jun 15, 2015 at 17:57

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