Cubes of a cubic

Because $\alpha$ is a root, we have $\alpha^3-3\alpha^2+5\alpha-7=0$ or

$\alpha^3=3\alpha^2-5\alpha+7$ , similarly we get

$\beta^3=3\beta^2-5\beta+7$

$\gamma^3=3\gamma^2-5\gamma+7$

Adding these three equations, we get

$\alpha^3+\beta^3+\gamma^3=3(\alpha^2+\beta^2+\gamma^2)-5(\alpha+\beta+\gamma)+21$

By Vieta, we have $\alpha+\beta+\gamma=3$ and $\alpha\beta+\alpha\gamma+\beta\gamma=5$ Writing the sum of squares as a linear combination of the elementary symmetric sums, we have the desired sum as

$3((\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma))-5(\alpha+\beta+\gamma)+21=3(3^2-2(5))-5(3)+21=\boxed{3}$

For any cubic equation
$a{ x }^{ 3 }+b{ x }^{ 2 }+cx+d$ with roots alpha, beta & gamma
$\alpha+\beta+\gamma\quad=\quad-\frac { b }{ a }$
$\alpha\beta\gamma \quad=\quad-\frac { d }{ a }$
$\alpha\beta +\beta\gamma +\gamma\alpha \quad=\quad\frac { c }{ a }$
We also have the identity
${ a }^{ 3 }+b^{ 3 }+c^{ 3 }-3abc=\quad(a+b+c)(a^{ 2 }{ +b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)$
${ a }^{ 3 }+b^{ 3 }+c^{ 3 }-3abc=\quad(a+b+c)([a{ +b }+{ c] }^{ 2 }-2ab-2bc-2ca-ab-bc-ca)$ ${ a }^{ 3 }+b^{ 3 }+c^{ 3 }-3abc=\quad(a+b+c)([a{ +b }+{ c] }^{ 2 }-3[ab+bc+ca])$

Substituting the values gives us
${ \alpha }^{ 3 }+\beta ^{ 3 }+\gamma ^{ 3 }-3(-\frac { d }{ a } )=\quad (-\frac { b }{ a } )([-\frac { b }{ a } ]^{ 2 }\quad-3[\frac{ c }{ a } ])$
${ \alpha }^{ 3 }+\beta ^{ 3 }+\gamma ^{ 3 }=(3)(3^{ 2 }-3[5])+3(7)$
${ \alpha }^{ 3 }+\beta ^{ 3 }+\gamma ^{ 3 }=3(-6)+21$
${ \alpha }^{ 3 }+\beta ^{ 3 }+\gamma ^{ 3 }=3$

$\begin{aligned} x^3-3x^2+5x-7 & =0 \\ x^3 & =3x^2-5x+7 \\ \sum x^3 & = \sum(3x^2-5x+7) \\ & = 3\sum x^2- 5 \sum x +\sum 7 \\ & = 3(-1)-5(3)+21 \\ & = \boxed{3} \end{aligned}$

thats same as mine,,,very nice