Find real root of cubic equation $(x-1)^3 - (x-1) + 4 = 0$

Question

a) $$(x-1)^3 - (x-1) + 4 = 0$$ When I expanded, I got $$x^3 - 3x^2 + 2x + 4 = 0$$ however I do not know what to do next.

b) $$8x^3 - 2x + 4 = 0$$ After I factored, I got $$2(4x^3 - x + 2) = 0$$ however I do not know what to do next.

Answer 1

For the cubic equation of the form $t^3 +pt + q = 0$, if $\frac{q^2}{4}+\frac{p^3}{27}>0$, use the Cardano's formula below for the analytic solution of the real root,

$$t= \left(-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\right)^{1/3}+\left(-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\right)^{1/3}\tag{1}$$

a) For $(x-1)^3 - (x-1) + 4 = 0$, we have $p=-1$ and $q=4$. Then, the real root is

$$x=1+\left( -2+\frac13\sqrt{\frac{107}{3}}\right)^{1/3}+\left( -2-\frac13\sqrt{\frac{107}{3}}\right)^{1/3}$$

b) For $8x^3 - 2x + 4 = 0$, we have $p=-\frac14$ and $q=\frac12$. Plug $p$ and $q$ into (1) to get the real root.