What is the sum of the cube of the roots of $ x^3 + x^2 - 2x + 1=0$?

Question

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.

Answer 1

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 $-10$

Answer 2

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.

Answer 3

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.

Answer 4

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