The equation \(x^{3}+3 x^{2}-4=0\) clearly has " \(x=1\) " as a positive solution.
(The other two solutions are \(x=-2,\) and \(x=-2-\) a repeated root; however
negatives were viewed with suspicion in the sixteenth century, so this root
might well have been ignored.) Try to understand how the following sequence of
moves "finds the root \(x=1 "\) :
(i) substitute \(y=x+1\) to get a cubic equation in \(y\) with no term in \(y^{2}\);
(ii) imagine \(y=u+v\) and interpret the identity for
$$
(u+v)^{3}=u^{3}+3 u v(u+v)+v^{3}
$$
as your cubic equation in \(y\);
(iii) solve the simultaneous equations " \(3 u v=3 ", " u^{3}+v^{3}=2\) " (not
by guessing, but by substituting \(v=\frac{1}{u}\) from the first equation into
the second to get a quadratic equation in " \(u^{3}\) ", which you can then
solve for \(u^{3}\) before taking cube roots);
(iv) then find the corresponding value of \(v\), hence the value of \(y=u+v,\) and
hence the value of \(x\).
The simple method underlying Problem 135 is in fact completely general. Given
any cubic equation
$$
a x^{3}+b x^{2}+c x+d=0 \quad \text { (with } \left.a \neq 0\right)
$$
we can divide through by \(a\) to reduce this to
$$
x^{3}+p x^{2}+q x+r=0
$$
with leading coefficient \(=1 .\) Then we can substitute \(y=x+\frac{p}{3}\) and
reduce this to a cubic equation in \(y\)
$$
y^{3}-3\left(\frac{p}{3}\right)^{2} y+q
y+\left[r+2\left(\frac{p}{3}\right)^{3}-q\left(\frac{p}{3}\right)\right]=0
$$
which we can treat as having the form
$$
y^{3}-m y-n=0
$$
So we can set \(y=u+v\) (for some unknown \(u\) and \(v\) yet to be chosen), and
treat the last equation as an instance of the identity
$$
(u+v)^{3}-3 u v(u+v)-\left(u^{3}+v^{3}\right)=0
$$
which it will become if we simply choose \(u\) and \(v\) to solve the simultaneous
equations
$$
3 u v=m, \quad u^{3}+v^{3}=n .
$$
We can then solve these equations to find \(u,\) then \(v-\) and hence find
\(y=u+v\) and \(x=y-\frac{p}{3}\).