Write down the values of the coefficient $b$ and $d$ for which the graph of $f(x)$ with equation $f(x) = ax^3 + bx^2 + cx + d$ would be symmetric about the origin.
Because this is an odd function, it leads me to the bx^2+d=0 equation. I am not sure that I am on the right way to the solution.
Thank you!
You want that $f(-x)=-f(x)$ for any $x$. In particular, $f(-1)=-f(1)$, so $$ -a+b-c+d=-a-b-c-d $$ whence $b+d=0$. Similarly, you need $f(-0)=-f(0)$, so $$ d=-d $$ whence $d=0$.
The two conditions imply $b=d=0$. The function $f(x)=ax^3+cx$ is odd for any value of $a$ and $c$, so you're done.