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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

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    $\begingroup$ If you are asked to solve for the roots of $4t^3-t-1$, i.e. find which values of $t$ satisfy $4t^3-t-1=0$, then you can use cardano's formula but noone memorizes it. Otherwise, it will be very difficult to do by hand. $\endgroup$ – JMoravitz Sep 13 '17 at 18:26
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    $\begingroup$ @projectilemotion's point with his comments is that "$4t^3-t-1$" is not an equation as it has no equals sign, it is an expression. This is just how "$5$" is not an equation. This has no truth value associated to it. Equations are true or false. Expressions don't have any intrinsic truth value. On the other hand "$4t^3-t-1=0$" is a perfectly valid equation and there will be certain values of $t$ which make it true and other values of $t$ which make it false. Solving an equation is generally finding which values of the variables present will make the equation true. $\endgroup$ – JMoravitz Sep 13 '17 at 18:30
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A cubic equation. This one has one real root: $$ \frac16\,\sqrt [3]{27+3\,\sqrt {78}}+{\frac {1}{2 \sqrt [3]{27+3\, \sqrt {78}}}} $$

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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 $$t=\frac{1}{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(3 \sqrt{3}\right)\right)$$

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