1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Are you sure you have to find the sum of the cube of the roots? $\endgroup$ – flawr Sep 27 '15 at 16:51
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ A hint $\endgroup$ – A.Γ. Sep 27 '15 at 16:56
  • 2
    $\begingroup$ Lol 3 answers exactly same $\endgroup$ – Display name Sep 27 '15 at 17:07
2
$\begingroup$

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 $-10$

$\endgroup$
3
  • $\begingroup$ David, can you explain step by step? $\endgroup$ – Lucas Henrique Sep 27 '15 at 17:39
  • 1
    $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

Hint: use Newton's formulas and Vieta's formulas https://brilliant.org/wiki/newtons-identities/

$\endgroup$
2
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.