It's always fascinating to see what's going on in textbooks of the olden days, and National Treasure @mathsjem recently found a beauty of its type. Look at those whences! Check out the subjunctives! It thrills the heart, doesn't it?1

What caught my attention, though, was *evolution* - in this context, taking square or cube (or, presumably, higher) roots of an expression.

I know! There's a whole chapter on each of them. Looks complicated... pic.twitter.com/DWPH4e96Ip

— Jo Morgan (@mathsjem) January 3, 2018

I haven't studied this method in detail - it looks similar in flavour to finding square roots by long division, which International Legend @colinthemathmo has discussed, at least for numbers .

However, it got me thinking: how would I find the cube root of $c(x) = 27 + 108x + 90x^2 + 80x^3 - 60x^4 + 48x^5 - 8x^6$?

If we're allowed to assume that $c(x)$ is a perfect cube, then it's not tricky at all: it's a degree-6 polynomial, so its cube root must be quadratic; the constant term is clearly 3 and the $x^2$ coefficient is just as clearly -2.

That means $c(x) = \br{ 3 + bx - 2x^2}^3$, for some value of $b$.

Assuming $x$ is small enough that we can ignore $x^2$ terms and higher, expanding the bracket with the binomial expansion gives $c(x) \approx 27 + 27bx$, and $b=4$.

Expanding the whole thing properly gives the same answer - which I leave as an exercise.

As an alternative to assuming the form of the solution, it's also possible to use the general binomial expansion (at least for small $x$) to figure out the cube root of $c(x)$.

Suppose we write $c(x) = 27 + x\br{108 + 90x + 80x^2 + ...}$ and work out $\br{c(x)}^{\frac{1}{3}}$?

The general form for $\br{a+bx}^{\frac{1}{3}}$ is $a^{\frac{1}{3}} + \frac{1}{3}a^{-\frac{2}{3}}bx - \frac{1}{9}a^{-\frac{5}{3}}b^2x^2 + ...$

That looks a mess; however, $a$ behaves quite nicely here and simplifies the whole thing down to $3 + \frac{1}{27} bx - \frac{1}{3^{7}}b^2 x^2 + ...$

There's a bit of book-keeping to do here, because $b$ is a bit of a monster, but stick with it: $\frac{1}{27} bx = 4x + \frac{10}{3}x^2 + ...$, while $\frac{1}{3^7} b^2x^2 = \frac{108^2}{3^7} x^2 + ...$.

What's $\frac{108^2}{3^7}$? I don't know off the top of my head. I do know that $108 = 2^2 \times 3^3$, so $108^2 = 2^4 \times 3^6$ ...

*Whoosh*

"It's 11,664, because $(100+k)^2 = 10,000 + 200k + k^2$."

*Whoosh*

... which leaves us with $\frac{16}{3}$.

So, we have an answer -- assuming we only need to go up to the $x^2$ term -- of $3 + 4x + \br{\frac{10}{3}- \frac{16}{3}}x^2$, which works out to $3 + 4x - 2x^2$ again.

The only problem left is checking whether we really *have* gone far enough. One option is to cube the answer, but you don't really want to be doing that every time you extract a term.

Presumably, something in the original method given in Jo's Victorian textbook checks for a remainder, but I don't really feel like wading through it to figure it out. Perhaps you, dear reader, can enlighten us all?

- If one were so inclined, one might even shiver in ecstasy. [↩]