I need someone to check if my proof is correct. We haven't been provided solutions with problem sheets and professor doesn't answer his mails :-(
Let's write polynomial as a function $f(x)= a_3x^3+a_2x^2+a_1x+a_0$. So if this function has more than one real zero (let's say two different zero's) than its graph will intersect x-axis at certain point. To make another zero possible, graph of a function $f$ will have to change its monotony (from ascending to descending and vice-versa, idk if "monotony" is the correct term). For two real zero's at least one change of monotony is required, for 3 it's 2 and for $n$ zeros it's $n-1$. If we set $n=4$ then we need 3 changes of monotony. This change appears when derivative of a function changes its sign at one point. For that to be possible $f'(x)$ must be zero at that point. Since $f'(x)$ is quadratic polynomial it has at most two real solution thus cubic polynomial cannot have more than 3 real solutions.
(If zero of a function is not correct expression in English(I am not quite good at translating mathematical terms), it means for certain $x, f(x)=0$)