Help with homework of 3rd order equation I can't solve this equation :

$$t^3-4t^2-5=0$$

I think it should be solved by cover it to $(a+b)^3$, but I can't figure it out 
 A: $t^3-4t^2-5=0$ does not factor nicely, and by the Rational Zero Theorem, the possible rational zeros (by $\frac{P}{Q}, -5/1$) are $1, -1, 5, -5$. Through synthetic division (see image), we find that none of these possible rational zeros are actual rational zeros, and thus there are no rational zeros, and we must use a calculator to solve this.
In doing so, we find that the zeros, to three decimal places, are $x = 4.274, -0.137 \pm 1.073i$.

A: Starting with GAVD'answer, which generates the depressed cubic equation in $u$
$$u^3-\frac{16}{3}u - \frac{263}{27}=0$$ use Vieta's substitution with $p=-\frac{16}{3}$ and $q=- \frac{263}{27}$.
This leads to the sextic equation in $w$ $$w^6-\frac{263 }{27}w^3+\frac{4096}{729}=0$$ which is quadratic in $w^3$. The solutions are $$w^3=\frac{1}{54} \left(263\pm3 \sqrt{5865}\right)$$
Just have a look at the Wikipedia page for the remaining.
A: Let $u=t-\frac{4}{3}$. One then has $$(u+\frac{4}{3})^3-4(u+\frac{4}{3})^2-5 = 0$$
$$u^3-\frac{16}{3}u - \frac{263}{27}=0.$$
Now, let $x+y = u$, then $x^3+y^3 = (x+y)^3-3xy(x+y) = u^3 - 3xyu$.
One can set $xy = \frac{16}{9}$, $x^3+y^3= \frac{263}{27}$. One can solve to find $x$, $y$ then $u= x+y$.
