<s>[INST] Use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ I have this equation 
$$x^3-x^2-x+0.5=0$$
Now i have to use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ which will be the roots of the new equation obatained.
I know that i can use many other ways but how to solve by this method
I end up with this
$$y-y^{\frac {2}{3}}-y^{\frac {1}{3}}+0.5=0$$
I can't factor out $y^{\frac{1}{3}}$
Any help is appriciated
 [/INST] We have $$2x^3+1=2x^2+2x$$
Cubing both sides  $$8(x^3)^3+1^3+3(2x^3)(2x^3+1)=(2x^2+2x)^3=8(x^3)^2+8(x^3)+3(2x^2)(2x)(2x^2+2x)$$
Now  as $x^3=y, 2x^2+2x=2x^3+1=2y+1$
$$8y^3+1+6y(2y+1)=8y^2+8y+12y(2y+1)$$
$$\implies8y^3+y^2(12-8-24)+(\cdots)y+(\cdots)=0$$ whose roots are 
$$\alpha^3,\beta^3,\gamma^3$$
Using Vieta's formula $$\alpha^3+\beta^3+\gamma^3=-\dfrac{12-8-24}8$$
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