Solve for a
\left\{\begin{matrix}a=-\frac{x^{3}-3y+3}{3x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&y=1\text{ and }x=0\end{matrix}\right.
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\frac{1}{3}x^{3}+ax+1=y
Swap sides so that all variable terms are on the left hand side.
ax+1=y-\frac{1}{3}x^{3}
Subtract \frac{1}{3}x^{3} from both sides.
ax=y-\frac{1}{3}x^{3}-1
Subtract 1 from both sides.
xa=-\frac{x^{3}}{3}+y-1
The equation is in standard form.
\frac{xa}{x}=\frac{-\frac{x^{3}}{3}+y-1}{x}
Divide both sides by x.
a=\frac{-\frac{x^{3}}{3}+y-1}{x}
Dividing by x undoes the multiplication by x.
a=-\frac{x^{2}}{3}+\frac{y-1}{x}
Divide y-\frac{x^{3}}{3}-1 by x.
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