<s>[INST] 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 
 [/INST] 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 $u= x+y$.
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