# Formula for polynomial

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.

• 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. – Ethan Bolker May 22 '17 at 16:34
• Please see the edit. – mathnoob123 May 22 '17 at 16:37
• @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? – mathnoob123 May 22 '17 at 16:47

$$(\alpha+\beta+\gamma)^3=\alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-3\alpha\beta\gamma$$

• Okay perfect. That's what I was looking for. – mathnoob123 May 22 '17 at 17:00
• Can this formula work for quartic equations? – mathnoob123 May 22 '17 at 17:01
• @Faiq Raees Do you mean for a resolvente of a quartic equation? – Michael Rozenberg May 22 '17 at 17:08
• 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? – mathnoob123 May 22 '17 at 17:12
• @Faiq Raees This identity is always true because it's identity. There is a similar identity for $(\alpha+\beta+\gamma+\delta)^4$. – Michael Rozenberg May 22 '17 at 17:18

do you mean this here $$(a+b+c)^3=a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3$$ ?

• Yes. But how to express this in summation form? – mathnoob123 May 22 '17 at 16:43

There is a similar identity for four numbers $a$, $b$, $c$ and $d$: $$(a+b+c+d)^4=a^4+b^4+c^4+d^4+4(a+b+c+d)^2(ab+ac+ad+bc+bd+cd)-$$ $$-2(ab+ac+ad+bc+bd+cd)^2-4(a+b+c+d)(abc+abd+acd+bcd)+4abcd.$$