Notes: What is x^3+y^3+z^3, if x,y,z are the solutions of the equation x^3-2x^2+2x+17=0?

Wednesday, April 29, 2015

If x,y,z are the roots of the given equation, then
substituted within equation, they verify
it.

$\displaystyle{{x}}^{{3}}-{2}{{x}}^{{2}}+{2}{x}+{17}={0}$

$\displaystyle{{y}}^{{3}}-{2}{{y}}^{{2}}+{2}{y}+{17}={0}$

$\displaystyle{{z}}^{{3}}-{2}{{z}}^{{2}}+{2}{z}+{17}={0}$

We'll
add the equations
above:

$\displaystyle{{x}}^{{3}}-{2}{{x}}^{{2}}+{2}{x}+{17}+{{y}}^{{3}}-{2}{{y}}^{{2}}+{2}{y}+{17}+{{z}}^{{3}}-{2}{{z}}^{{2}}+{2}{z}+{17}={0}$

We'll
isolate to the left side the sum of cubes:

$\displaystyle{{x}}^{{3}}+{{y}}^{{3}}+{{z}}^{{3}}$
= 2x^2 - 2x - 17 + 2y^2 - 2y - 17 + 2z^2 - 2z - 17

$\displaystyle{{x}}^{{3}}+$
y^3 + z^3 = 2(x^2 + y^2 + z^2) - 2(x + y + z) - 3*17

We'll
use Viete's relations to determine the sum of the roots and the sum of the squares of
the roots.

$\displaystyle{x}+{y}+{z}=-\frac{{-{2}}}{{1}}=$
2

$\displaystyle{{x}}^{{2}}+{{y}}^{{2}}+{{z}}^{{2}}={{\left({x}+{y}+{z}\right)}}^{{2}}-{2}{\left({x}{y}+{x}{z}+\right.}$
yz)

We'll use Viete's relations again to calculate the sum
of the products of two roots:

$\displaystyle{x}{y}+{x}{z}+{y}{z}=\frac{{2}}{{1}}=$
2

$\displaystyle{{x}}^{{2}}+{{y}}^{{2}}+{{z}}^{{2}}={{\left({2}\right)}}^{{2}}-{2}\cdot{2}={4}-{4}=$
0

$\displaystyle{{x}}^{{3}}+{{y}}^{{3}}+{{z}}^{{3}}={2}\cdot{\left({0}\right)}-{2}\cdot{\left({2}\right)}-$
3*17

$\displaystyle{{x}}^{{3}}+{{y}}^{{3}}+{{z}}^{{3}}=-{4}-$
51

$\displaystyle{{x}}^{{3}}+{{y}}^{{3}}+{{z}}^{{3}}=$
-55

Therefore, the sum of the cubes of
the roots of equation is:

$\displaystyle{{x}}^{{3}}$
+ y^3 + z^3 = -55

Notes