Vamos a calcular el cubo de los binomios en cada uno de los siguientes casos, utilizando la fórmula del cubo de un binomio [tex]\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)[/tex]. Sigamos el procedimiento:
a. Calcular [tex]\((a + 2)^3\)[/tex]:
[tex]\[
(a + 2)^3 = a^3 + 3a^2(2) + 3a(2^2) + 2^3
\][/tex]
[tex]\[
= a^3 + 6a^2 + 12a + 8
\][/tex]
b. Calcular [tex]\((a \cdot 4)^3 = (4a)^3\)[/tex]:
[tex]\[
(4a)^3 = (4)^3(a)^3
\][/tex]
[tex]\[
= 64a^3
\][/tex]
c. Calcular [tex]\(\left(m - \frac{2}{7}\right)^3\)[/tex]:
[tex]\[
\left(m - \frac{2}{7}\right)^3 = m^3 + 3m^2\left(-\frac{2}{7}\right) + 3m\left(-\frac{2}{7}\right)^2 + \left(-\frac{2}{7}\right)^3
\][/tex]
[tex]\[
= m^3 - \frac{6m^2}{7} + \frac{12m}{49} - \frac{8}{343}
\][/tex]
d. Calcular [tex]\(\left(m + \frac{5}{4}\right)^3\)[/tex]:
[tex]\[
\left(m + \frac{5}{4}\right)^3 = m^3 + 3m^2\left(\frac{5}{4}\right) + 3m\left(\frac{5}{4}\right)^2 + \left(\frac{5}{4}\right)^3
\][/tex]
[tex]\[
= m^3 + \frac{15m^2}{4} + \frac{75m}{16} + \frac{125}{64}
\][/tex]
e. Calcular [tex]\(\left(\frac{2}{3} + x\right)^3\)[/tex]:
[tex]\[
\left(\frac{2}{3} + x\right)^3 = x^3 + 3x^2\left(\frac{2}{3}\right) + 3x\left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3
\][/tex]
[tex]\[
= x^3 + \frac{6x^2}{3} + \frac{12x}{9} + \frac{8}{27}
\][/tex]
[tex]\[
= x^3 + 2x^2 + \frac{4x}{3} + \frac{8}{27}
\][/tex]
f. Calcular [tex]\(\left(n - \frac{2}{7}\right)^3\)[/tex]:
[tex]\[
\left(n - \frac{2}{7}\right)^3 = n^3 + 3n^2\left(-\frac{2}{7}\right) + 3n\left(-\frac{2}{7}\right)^2 + \left(-\frac{2}{7}\right)^3
\][/tex]
[tex]\[
= n^3 - \frac{6n^2}{7} + \frac{12n}{49} - \frac{8}{343}
\][/tex]
Por lo tanto, los resultados son:
[tex]\[
\begin{array}{ll}
\text{a. } a^3 + 6a^2 + 12a + 8 \\
\text{b. } 64a^3 \\
\text{c. } m^3 - \frac{6m^2}{7} + \frac{12m}{49} - \frac{8}{343} \\
\text{d. } m^3 + \frac{15m^2}{4} + \frac{75m}{16} + \frac{125}{64} \\
\text{e. } x^3 + 2x^2 + \frac{4x}{3} + \frac{8}{27} \\
\text{f. } n^3 - \frac{6n^2}{7} + \frac{12n}{49} - \frac{8}{343}
\end{array}
\][/tex]
