I was looking for a solution to the following equation:
$$x^4 - x^3 = C$$
where $x$ is real and $C$ is a parameter.
Wolfram Alpha returned a very complex answer. I am looking for a simpler answer, which is approximately correct when $C$ is sufficiently large (i.e., it approaches the optimal solution when $C\to\infty$).
How can I find such a solution for this equation in particular, and for other polynomial equations in general?
Consider that $$x^4-x^3=x^4-x^3+O\left(\frac{1}{x^4}\right)\qquad\qquad \color{red}{\large (!!)}$$ and use series reversion and, in the real domain, you should get (using $d=\sqrt[4]{C}$) $$x=d+\frac{1}{4}+\frac{3}{32 d}+\frac{1}{32 d^2}+\frac{15}{2048 d^3}-\frac{77}{65536 d^5}-\frac{3}{4096 d^6}+O\left(\frac{1}{d^7}\right)$$ Trying for a small number $C=100$, this truncated formula will give $$x=\frac{16588752+66155123 \sqrt{10}}{65536000}=\color{red}{3.4452761}77$$ while the exact solution is $\color{red}{3.445276105}$.
Using this formula $$x^4-x^3-C=\frac{1989}{2097152 C}+O\left(\frac{1}{C^{3/2}}\right)\sim \frac 1 {1000 C}$$
When $C$ is sufficiently large, then $|x|>>1$ holds because following graph.
So $C=x^4-x^3=x^3(x-1)\sim x^4$. Asymptotic answer will be $x=\pm^4\sqrt C$.
+) Or more accurate answer.
$(x-\frac14)^4=x^4-x^3+\frac38x^2-\frac1{16}x+\frac1{256}\sim x^4-x^3$, so we can approximate $x^4-x^3$ as $(x-\frac14)^4$.
So the answer is $x=\frac14\pm^4\sqrt C$.
How in general ...?
Factor out the big part. In this case, the power of $x$ of largest degree. \begin{align*} C &= x^4 - x^3 \\ &= x^4 \left( 1 - \frac{1}{x} \right) \text{.} \end{align*} Here, we see the approximate solutions $x \in \{\sqrt[4]{C}, \mathrm{i}\sqrt[4]{C}, -\sqrt[4]{C}, -\mathrm{i}\sqrt[4]{C} \}$. (If you do not want complex approximate solutions, ignore the complex approximate solutions; that still leaves two real approximate solutions.) This approximation improves "as fast as" $-1/x$ becomes negligible in your application.
More generally, take the polynomial $P(x)$, of degree $n$, written in the form $$ P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0 \text{.} $$ Then $$ a_n x^n \left( \frac{a_{n-1}}{a_n x} + \frac{a_{n-2}}{a_n x^2} + \cdots + \frac{a_1}{a_n x^{n-1}} \right) = -a_0 \text{.} $$ Approximate solutions are the $n$ roots of $$ x^n = \frac{-a_0}{a_n} \text{.} $$ (And again, ignore the complex approximate solutions if they are not of use to you. If so, if $n$ is even, you get positive and negative $n^{\text{th}}$ roots and if $n$ is odd, only the root whose sign matches that of $-a_0/a_n$.) If we are interested in large $x$, this approximation improves at least as fast as $1/x$, although large coefficients in the parenthesized term can delay this a bit (until the power of $x$ in the relevant term is larger than the coefficient, so starts to overwhelm it).
Can we do better if there is useful structure? Sure, but such methods depend on the structure and you have given no hint what kind of structure to expect.
Let $C\to t$ and $Q^4=t$, then $Q=\left\{-\sqrt[4]{t},-i \sqrt[4]{t},i \sqrt[4]{t},\sqrt[4]{t}\right\}$.
First approximation $x=\dfrac{2 Q (2 Q-1)}{4 Q-3}$.
More exact approximation $x=\dfrac{Q \left(1024 Q^5-2816 Q^4+3168 Q^3-1776 Q^2+481 Q-48\right)}{4 (2 Q-1)^2 (4 Q-3) \left(16 Q^2-20 Q+9\right)}$.
More-more exact approximation $x=\frac{Q \left(4398046511104 Q^{21}-51677046505472 Q^{20}+291095703453696 Q^{19}-1044845284032512 Q^{18}+2679340185681920 Q^{17}-5216549166120960 Q^{16}+7999588518592512 Q^{15}-9892945964564480 Q^{14}+10019888069345280 Q^{13}-8393581748289536 Q^{12}+5847885318586368 Q^{11}-3395429015715840 Q^{10}+1640930164996096 Q^9-657039601832960 Q^8+216160980335232 Q^7-57679294645056 Q^6+12246042446563 Q^5-2011062174240 Q^4+244701755352 Q^3-20572244352 Q^2+1050589440 Q-23887872\right)}{16 (2 Q-1)^2 (4 Q-3) \left(16 Q^2-20 Q+9\right) \left(1024 Q^5-2816 Q^4+3168 Q^3-1776 Q^2+481 Q-48\right)^2 \left(1024 Q^6-3584 Q^5+5472 Q^4-4656 Q^3+2341 Q^2-660 Q+81\right)}$.
Verify code for Wolfram CAS:
t=RandomInteger[{-1000000,1000000}];
P=x^4-x^3-t;
Print["Equation: ",P,"=0"];
Print["Solution by CAS:"];
Print[NSolve[P==0,x,16]];
Q={-t^(1/4),-I t^(1/4),I t^(1/4),t^(1/4)};
Print["Solution by formula:"];
Print["x = ",N[(Q (-23887872+1050589440 Q-20572244352 Q^2+244701755352 Q^3-2011062174240 Q^4+12246042446563 Q^5-57679294645056 Q^6+216160980335232 Q^7-657039601832960 Q^8+1640930164996096 Q^9-3395429015715840 Q^10+5847885318586368 Q^11-8393581748289536 Q^12+10019888069345280 Q^13-9892945964564480 Q^14+7999588518592512 Q^15-5216549166120960 Q^16+2679340185681920 Q^17-1044845284032512 Q^18+291095703453696 Q^19-51677046505472 Q^20+4398046511104 Q^21))/(16 (-1+2 Q)^2 (-3+4 Q) (9-20 Q+16 Q^2) (-48+481 Q-1776 Q^2+3168 Q^3-2816 Q^4+1024 Q^5)^2 (81-660 Q+2341 Q^2-4656 Q^3+5472 Q^4-3584 Q^5+1024 Q^6)),16]//Sort]
