Solving for the value of $t$ in $4t^3-t-1=0$

Question

Solve for $t$: $$4t^3-t-1=0$$

I couldn't use polynomial. I tried this:

$t(4t^2-1)-1=0$ but then again it doesn't make sense to solve $t$ from that

Answer 1

A cubic equation. This one has one real root: $$ \frac16\,\sqrt [3]{27+3\,\sqrt {78}}+{\frac {1}{2 \sqrt [3]{27+3\, \sqrt {78}}}} $$

Answer 2

When you have a depressed cubic equation $t^3+pt+q=0$, a good idea is to use Vieta's substitution $t=w-\frac p {3w}$

$$t^3+pt+q=0\implies w^3+q-\frac{p^3}{27w^3}=0\implies w^6+qw^3-\frac{p^3}{27}=0$$ which is a quadratic equation in $w^3$.

In your case $p=q=-\frac 14$ which gives $$w^3=\frac{9+\sqrt{78}}{72}\implies t=\frac{\sqrt[3]{9+\sqrt{78}}}{2\ 3^{2/3}}+\frac{1}{2 \sqrt[3]{3 \left(9+\sqrt{78}\right)}}$$

If you want to look fancy, you also could use the hyperbolic solution for one real root and get $$t=\frac{1}{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(3 \sqrt{3}\right)\right)$$