<s>[INST] What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m$, determine the length of $AC$.(Answer$:\sqrt[3]{2})$

My progress:
$ AF = h\\
a=m+1\\
h^2 =m.FC = m.1 = m\\
\triangle AFC: AC^2 =b= h^2+FC^2 \implies b = \sqrt{m+1}\\
\triangle ABC: a^2 = b^2+ c^2 \implies (m+1)^2=c^2+(\sqrt{m+1})^2\\
\therefore c = \sqrt{m^2+m}\\
\triangle BAD: BD^2=AD^2+c^2\implies 1 = (\sqrt{m+1}-1)^2+m^2+m\\
\therefore: m^2+2m-2\sqrt{m+1}+1=0$
...????

 [/INST] Following on from $m^2+2m-2\sqrt{m+1}+1=0$, you already have $b=\sqrt{m+1}$, so substitute this in to get $b^4-2b=0 \implies b=\sqrt[3]2$.
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