Gazeta Matematica 6-7-8/2016, Problem 27248

Problem:
Solve in real numbers the equation: $$\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$\begin{array}{lll} (x^6-1)-(3x^5-5x^3+3x)&=&x^6-3x^5+5x^3-3x-1\\&=&(x^2-x-1)(x^4-2x^3-x^2+2x+1)\\&=&(x^2-x-1)(x^2-x-1)^2\\&=&(x^2-x-1)^3. \end{array}$$ Then, from the identity $(a-b)^3=a^3-b^3-3ab(a-b)$, the given equation can be written as
$$(x^6-1)-(3x^5-5x^3+3x)-3\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=(x^2-x-1)^3,$$ i.e. $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=0.$$ Since $\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1$, then the given equation becomes $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot(x^2-x-1)=0,$$ i.e. $$(x^6-1)(3x^5-5x^3+3x)(x^2-x-1)=0.$$ Finally, we have $$(x-1)(x^2+x+1)(x+1)(x^2-x+1)x(3x^4-5x^2+3)(x^2-x-1)=0.$$
Solving each equation, we get $x \in \left\{-1,0,1,\dfrac{1-\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right\}$.