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There is a formula which relate the roots: $$(\sum \alpha)^2=\sum \alpha^2-2\sum \alpha\beta$$

However I have kind of forgotten the formula which relates the $\sum \alpha^3$. (I think it's only used for cubic equations)

The formula kind of look like this $$(\sum \alpha)^3=\sum \alpha^3+3\sum \alpha\sum\alpha\beta+3\sum \alpha\beta\gamma$$ (This I think is wrong because I used it and got a wrong answer)

Can somebody please provide the formula?(I tried searching on Google but couldn't find it)

P.S. $\alpha,\beta,\gamma$ are roots of a general polynomial equation.

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  • $\begingroup$ It's not clear what "formula" you're looking for. A place to start reading is here: en.wikipedia.org/wiki/Elementary_symmetric_polynomial . There's a discussion of the relation between the roots and coefficients of a polynomial. $\endgroup$ – Ethan Bolker May 22 '17 at 16:34
  • $\begingroup$ Please see the edit. $\endgroup$ – mathnoob123 May 22 '17 at 16:37
  • $\begingroup$ @YvesDaoust Can you formula be altered to contain $\sum ab$ or $\sum a$ and not $\sum a^2b$ or $\sum ab^2$ since when dealing with polynomial its very easy to obtain those terms? $\endgroup$ – mathnoob123 May 22 '17 at 16:47
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$$(\alpha+\beta+\gamma)^3=\alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-3\alpha\beta\gamma$$

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  • $\begingroup$ Okay perfect. That's what I was looking for. $\endgroup$ – mathnoob123 May 22 '17 at 17:00
  • $\begingroup$ Can this formula work for quartic equations? $\endgroup$ – mathnoob123 May 22 '17 at 17:01
  • $\begingroup$ @Faiq Raees Do you mean for a resolvente of a quartic equation? $\endgroup$ – Michael Rozenberg May 22 '17 at 17:08
  • $\begingroup$ I have been using these formulas to work out the coefficient of polynomials, (if the sum of roots, roots squared etc.) is provided. So if i were to plugin the appropriate sums(that corresponds to the roots of a quartic equation) will this formula hold true? $\endgroup$ – mathnoob123 May 22 '17 at 17:12
  • $\begingroup$ @Faiq Raees This identity is always true because it's identity. There is a similar identity for $(\alpha+\beta+\gamma+\delta)^4$. $\endgroup$ – Michael Rozenberg May 22 '17 at 17:18
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do you mean this here $$(a+b+c)^3=a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3$$ ?

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  • $\begingroup$ Yes. But how to express this in summation form? $\endgroup$ – mathnoob123 May 22 '17 at 16:43
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There is a similar identity for four numbers $a$, $b$, $c$ and $d$: $$(a+b+c+d)^4=a^4+b^4+c^4+d^4+4(a+b+c+d)^2(ab+ac+ad+bc+bd+cd)-$$ $$-2(ab+ac+ad+bc+bd+cd)^2-4(a+b+c+d)(abc+abd+acd+bcd)+4abcd.$$

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