Use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$

Question

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

Answer 1

Hint: $$y+0.5=y^{\frac{2}{3}}+y^{\frac{1}{3}}$$ Now cube both sides to get a cubic equation in $y$. $$\Longrightarrow y^3+1.5 y^2+ 0.75 y+0.125 = y^2+y+3y(y^{\frac{2}{3}}+y^{\frac{1}{3}})$$

Now use the first equation again in it.

Answer 2

Hint : Use $$(y +0.5)^3 =y^2 +y+3y^{2 \over 3}.y^{1 \over 3}(y^{2 \over 3}+y^{1 \over 3})$$ Again use

$$ y^{2 \over 3}+y^{1 \over 3}=y+0.5$$ We got $$(y+0.5)^3=y^2+y+3y(y+0.5)$$ Solve it into cubic equations . And then you know the formula for sum of roots.

Answer 3

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 $$\alpha^3+\beta^3+\gamma^3=-\dfrac{12-8-24}8$$