2015/063) What are the real roots of $x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0$

The equation is very close to $(x-1)^6$ except coefficient of $x^3$ which is $-30 x^3$ instead of $-20 x^3$

so equation is

$(x-1)^6 - 10x^3 = 0$

let $\sqrt[3]10 = t$

so we get $(x-1)^6 - (tx)^3 = 0$

x is not zero

so divide by $x^3$
$(\dfrac{(x-1)^2}{x})^6 = t$

OR $(x - 1)^2 - \sqrt[3]{10}x  = 0$
ie $(x^2 - (2+ \sqrt[3]{10})x +1   = 0$
ie $x= \dfrac{1}{2}(2+\sqrt[3]{10})\pm\sqrt{4+4\sqrt[3]{10}+\sqrt[3]{100}-4}$
or  $x= \dfrac{1}{2}(2+\sqrt[3]{10})\pm\sqrt{4\sqrt[3]{10}+\sqrt[3]{100}}$