Proving roots of a function cannot all be real

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on. We have $$ \begin{align} f(x)&=x^6+ax^3+bx^2+cx+d\\ f'(x)&=6x^5+3ax^2+2bx+c\\ f''(x)&=30x^4+6ax+2b\\ f'''(x)&=120x^3+6a\\ \end{align} $$ Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$). Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$\text{Sum of the roots}=-\sum \alpha = 0.$$

Squaring, $$\sum \alpha^2 + 2 \sum \alpha \beta = \sum\alpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .