137. Find a number that has the property that when it is multiplied by 12, and added to its cube, it equals six times its square, increased by thirty-five (solved elementarily).
137. Equation:

$$
x^{3}+12 x=6 x^{2}+35
$$

Bhaskara's solutions:

$$
x^{3}+12 x-6 x^{2}=35
$$

Subtract 8 from both sides:

$$
\begin{aligned}
& x^{3}-6 x^{2}+12 x-8=27 \\
& (x-2)^{3}=27 \\
& x-2=3 \text { and } x=5
\end{aligned}
$$

To find the other two roots (which Bhaskara does not provide), we will solve this equation in a slightly different (but also elementary) way:

$$
\begin{aligned}
& x^{3}-6 x^{2}+12 x-35=0 \\
& x^{3}-5 x^{2}-x^{2}+5 x+7 x-35=0 \\
& x^{2}(x-5)-x(x-5)+7(x-5)=0 \\
& (x-5)\left(x^{2}-x+7\right)=0
\end{aligned}
$$

from which either $x-5=0$, and thus $x_{1}=5$, or $x^{2}-x+7=0$, from which it is easy to find $x_{2}$ and $x_{3}$.