Can the following expressions be factored into the sum of cubes of first-degree expressions:

(a.) $189 x^{3}+129 x^{2}+183 x+19$

(b.) $x^{3}+69 x^{2}+87 x+167 ?$
$(p x+q)^{3}+(r x+s)^{3}$ should be solved in this form, but it is sufficient, and somewhat easier, to look for the solution in the form $t(x+u)^{3}+$ $v(x+w)^{3}$. Then $p=\sqrt[3]{t}, q=u \sqrt[3]{t}, r=\sqrt[3]{v}, s=w \sqrt[3]{v}$. If we expand the latter expression term by term, we arrive at the following system of equations:

$$
\begin{aligned}
& \text { a) } t+v \quad=189, \quad \text { b) } t \quad+v=1, \\
& t u+v w=43, \quad t u+v w=23, \\
& t u^{2}+v w^{3}=61, \quad t u^{2}+v w^{3}=29, \\
& t u^{3}+v w^{3}=19 ; \quad t u^{3}+v w^{3}=167 .
\end{aligned}
$$

The solution of these systems of equations can be reduced to a quadratic, for example, by the method of the second solution of problem 65 /a (1. 25. o.). Multiply the second and third equations by $u+w$, then using our equations we get that

$$
\text { a) } \begin{aligned}
& 43(u+w)=61+189 u w \text { b) } \quad \begin{array}{l}
23(u+w)=29+u w \\
61(u+w)=19+43 u w
\end{array} \\
& 29(u+w)=167+23 u w
\end{aligned}
$$

From here

$$
u+w=\frac{1}{10}, \quad u w=-\frac{3}{10} \quad \text { or } \quad u+w=1, \quad u w=-6
$$

Thus, $u$ and $w$ are the two solutions of the following quadratic equation:

$$
10 z^{2}-z-3=0 \quad \text { or } \quad z^{2}-z-6=0
$$

so

$$
u, \quad w=-\frac{1}{2}, \quad \frac{3}{5} \quad \text { or } \quad \mathrm{u}, \quad \mathrm{w}=3, \quad-2
$$

and from the first two equations of the systems of equations, for example,

$$
t, \quad v=64, \quad 125=4^{3}, \quad 5^{3} \quad \text { or } \quad \mathrm{t}, \quad \mathrm{v}=5, \quad-4
$$

Thus, the sought decompositions are:

$$
\begin{gathered}
189 x^{3}+129 x^{2}+183 x+19=(4 x-2)^{3}+(5 x+3)^{3} \\
x^{3}+69 x^{2}+87 x+167=5(x+3)^{3}-4(x-2)^{3}
\end{gathered}
$$