# Euler's Derivation of the Cubic Formula

In his 1770 texbook, Elements of Algebra Euler gave a solution to the depressed equation x^3 = mx + n.

Assuming x = \sqrt[3]{p} + \sqrt[3]{q},  Euler cubed both sides:

 x^3 = p + 3\sqrt[3]{p^2q} + 3\sqrt[3]{pq^2} + q = 3\sqrt[3]{pq}(\sqrt[3]{p} + \sqrt[3]{q}) + (p + q) = 3\sqrt[3]{pq}x + (p + q).

The resulting equation obviously has the same structure as the original depressed cubic equation. This suggests letting 3\sqrt[3]{pq} = m  and p + q = n; and subsequently finding p   and q  and then also x = \sqrt[3]{p} + \sqrt[3]{q}.

So this is what Euler did. 4pq = 4m^3/27  and from p + q = n   p^2 + 2pq + q^2 = n^2. Combining these yields

(p^2 +2pq +q^2) - 4pq = n^2 - \frac{4m^3}{27}

which simplifies to

(p - q)^2 = n^2 - \frac{4m^3}{27}.

Thus

p - q = \sqrt{n^2 - \frac{4m^3}{27}}.

Adding and subtracting p + q = n,  gives

 2p = n + \sqrt{n^2 - \frac{m^3}{27}} 2q = n - \sqrt{n^2 - \frac{m^3}{27}}

So that the solution of the original cubic is

x = \sqrt[3]{\frac{n}{2} + \sqrt{\frac{n^2}{4} - \frac{m^3}{27}} } + \sqrt[3]{\frac{n}{2} - \sqrt{\frac{n^2}{4} - \frac{m^3}{27}} }.

## Reference

1. W. Dunham, Euler: The Master of Us All, MAA, 1999, pp. 83-84