Let $k$ be a field s.t. $\mathop{\mathrm{char}}k\neq2,3$. Let $d\geq 1$, $f\in k[x]$ be a polynomial of degree $4d$, $g\in k[x]$ be a polynomial of degree $6d$ with $\gcd(f,g)=1$. So $f^3$ and $g^2$ are both polynomials of degree $12d$.
Can we pick $f,g$ s.t. $\deg(f^3 +g^2)<d$?
This problem is in a way a generalization of this question, in particular, we can't achieve this with $d=1$, since $f^3 +g^2$ cann't be zero or non-zero constant.
We can consider the coefficients of $f$ as a vector $v_f$ of dimension $4d+1$, similarly the coefficients of $g$ is a vector $v_g$ of dimension $6d+1$. Then the coefficient of $h=f^3+g^2$ is a vector $v_h$ of dimension $12d+1$, and each entry of $v_h$ is a function on $v_f$ and $v_g$. The condition $\deg h<n$ imples that top $12d+1-n$ entries of $v_h$ are zero.
If $12d+1-n>(4d+1)+(6d+1)=10d+2$ (equivalently $n<2d-1$), we have that the number of constrains is greater than the number of free variables, then it is quite possible there are no non-trivial solutions.
That would contradict the Mason-Stothers theorem. Indeed, as soon as the theorem is applicable, we have $\mathop{\mathrm{deg}}\mathop{\mathrm{rad}} (f^3g^2(f ^3+g^2))\geq 12d+1$, wherefore $\mathop{\mathrm{deg}} (f ^3+g^2)\geq 2d+1$.
The theorem may be inapplicable only if $f’=g’=0$, I.e. when $f$ and $g$ are both $p$th powers of some polynomials (over an extension of $k$): where $p=\mathop{\mathrm{char}} k$. Then one may apply the same argument to their $p$th degree roots.