I was reading An Imaginary Tale: The Story of the Square Root of Negative One and it begins with a derivation of the general solution for depressed cubics.
It begins with the format for depressed cubics:
$$x^3+px=q$$
Then says that $x$ can be written where $x=u+v$, which inserted into the original question gives
$$u^3+v^3+(3uv+p)(u+v)=q$$
It then says, "This single, rather complicated-looking equation, can be rewritten as two individually less complicated statements:"
$$3uv+p=0$$
and $$u^3+v^3=q$$
My question is why are these two equations the same as the one equation above?
$$u^3+v^3+(3uv+p)(u+v)=q$$$$u^3+v^3=(u+v)(u^2-uv+v^2)$$$$(u+v)(u^2-uv+v^2)+(3uv+p)(u+v)=q$$$$(u+v)(u^2-uv+v^2+3uv+p)=(u+v)(u^2+v^2+2uv+p)$$$$u^2+2uv+(v^2+p)=(u-\frac{-2v+\sqrt{-4p}}{2})(u-\frac{-2v-\sqrt{-4p}}{2})$$$$(u+v)(u-\frac{-2v+\sqrt{-4p}}{2})(u-\frac{-2v-\sqrt{-4p}}{2})=q$$
Does that help?