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I know there are roots, because if we assume the equation as a function and give -3 and 1 as $x$:

$$ (-3)^3 + (-3)^2 - 2(-3) + 1 <0 $$

$$ 1^3 + 1^2 - 2(1) + 1 > 0 $$

It must have a root between $[-3,1]$. However, the root is very hard and it appeared on a high school test. How can I solve it simply?

The given options were -10, -5, 0 , 5 and 10. Note: we didn't even learn the cube root formula. The test had just logic problems and I didn't use any calculus or complicated stuff. So there must be an easier way without using cube root concepts or formulas.

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    $\begingroup$ Are you sure you have to find the sum of the cube of the roots? $\endgroup$ – flawr Sep 27 '15 at 16:51
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    $\begingroup$ A quick search provided some useful information I think: math.stackexchange.com/questions/30491/sum-of-cubed-roots $\endgroup$ – flawr Sep 27 '15 at 16:53
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    $\begingroup$ A hint $\endgroup$ – A.Γ. Sep 27 '15 at 16:56
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    $\begingroup$ Lol 3 answers exactly same $\endgroup$ – Display name Sep 27 '15 at 17:07
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using the information in the equation $$ \sum_{i=1}^3 r_i^3 = -\sum_{i=1}^3 (r_i^2 - 2r_i + 1) \\ = -5 -\sum_{i=1}^3 r_i^2 $$ also, from the well-known expressions giving the elementary symmetric functions of the roots in terms of the coefficients, $$ \sum_{i=1}^3 r_i^2 = (\sum_{i=1}^3 r_i)^2 - 2(r_1r_2+r_2r_3+r_3r_1) \\ = 5 $$ so the sum of cubes of the roots is $-10$

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  • $\begingroup$ David, can you explain step by step? $\endgroup$ – Lucas Henrique Sep 27 '15 at 17:39
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    $\begingroup$ if $r$ is a root, then $r^3 = -r^2+2r-1$. a cubic has three roots (in an algebraically closed field), so we can sum this relation over the roots $r_1,r_2,r_3$. the constant term $-1$ gives $-3$. the sum of the roots is minus the coefficient of $x^2$ divided by the coefficient of $x^3$, so the $2r$ term gives a sum of $-2$. the $r^2$ term is obtained by the method shown, where the sum of the products of the roots in pairs is the ratio of the coefficient of $x$ to that of $x^3$. check out one of the references in other answers to "relations between roots" etc. $\endgroup$ – David Holden Sep 27 '15 at 17:59
  • $\begingroup$ Perfect! You did it. $\endgroup$ – Lucas Henrique Sep 27 '15 at 18:00
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You know,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$

$a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc$

Now I hope you know the relation between roots.

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Suppose $a,b,c\in\mathbb{C}$ are the roots to your equation. There is this algebraic identity you should know: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ Do some manipulations to get this form: $$a^3+b^3+c^3=3abc+(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)$$

Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and products of its roots.

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  • $\begingroup$ I'm curious: is there any relationship with $ (x-1)^2 $ ? $\endgroup$ – Lucas Henrique Sep 27 '15 at 17:08
  • $\begingroup$ @Lucas maybe. i don't see one $\endgroup$ – najayaz Sep 27 '15 at 17:11
  • $\begingroup$ Take a look at the note. $\endgroup$ – Lucas Henrique Sep 27 '15 at 17:13
  • $\begingroup$ @LucasHenrique this is the easiest way I can think of. It does not involve any calculus. The algebraic identity seems simple enough and as for Vieta's formulas, I am confident that someone your age can understand that stuff. $\endgroup$ – najayaz Sep 27 '15 at 17:19
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Hint: use Newton's formulas and Vieta's formulas https://brilliant.org/wiki/newtons-identities/

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    $\begingroup$ I think the link is to newton $\endgroup$ – Display name Sep 27 '15 at 17:03
  • $\begingroup$ Thnx its ok now $\endgroup$ – Display name Sep 27 '15 at 17:05

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