Question

Productos notables
Complete el desarrollo de cada binomio
A) (t^2 + 2t)^3 =
B) ( 3x/4 - 4/3y)^3=
C) ( a^2 - b4)^3 =
D) ( 2a + 3b)^3 =
E) ( 2a/5 + 5/2b) ^3 =
F) ( 2/5a + 5/2b)^3 =

Ayudenme porfa es urgente !

Answer 1

El desarrollo de cada binomio se obtiene por el uso de la fórmula:

(a  ±  b)³  =  a³  ±  3a²b  +  3ab²  ±  b³

Explicación paso a paso:

Aplicaremos la fórmula general del binomio al cubo en cada uno de los productos notables solicitados:

$\bold{A)\quad(t^{2}+2t)^{3}}$

$(t^{2}+2t)^{3}=(t^{2})^{3}+3(t^{2})^{2}(2t)+3(t^{2})(2t)^{2}+(2t)^{3} \qquad \Rightarrow\\\\\bold{(t^{2}+2t)^{3}=t^{6}+6t^{5}+12t^{4}+8t^{3}}$

$\bold{B)\quad(\frac{3x}{4}-\frac{4}{3y})^{3}}$

$(\frac{3x}{4}-\frac{4}{3y})^{3}=(\frac{3x}{4})^{3}-3(\frac{3x}{4})^{2}(\frac{4}{3y})+3(\frac{3x}{4})(\frac{4}{3y})^{2}-(\frac{4}{3y})^{3} \qquad \Rightarrow\\\\\bold{(\frac{3x}{4}-\frac{4}{3y})^{3}=\frac{27x^{3}}{64}-\frac{9x^{2}}{4y}+\frac{4x}{y^{2}}-\frac{64}{27y^{3}}}$

$\bold{C)\quad(a^{2}-b^{4})^{3}}$

$(a^{2}-b^{4})^{3}=(a^{2})^{3}-3(a^{2})^{2}(b^{4})+3(a^{2})(b^{4})^{2}-(b^{4})^{3} \qquad \Rightarrow\\\\\bold{(a^{2}-b^{4})^{3}=a^{6}-3a^{4}b^{4}+3a^{2}b^{8}-b^{12}}$

$\bold{D)\quad(2a+3b)^{3}}$

$(2a+3b)^{3}=(2a)^{3}+3(2a)^{2}(3b)+3(2a)(3b)^{2}+(3b)^{3} \qquad \Rightarrow\\\\\bold{(2a+3b)^{3}=8a^{3}+36a^{2}b+54ab^{2}+27b^{3}}$

$\bold{E)\quad(\frac{2a}{5}+\frac{5}{2b})^{3}}$

$(\frac{2a}{5}+\frac{5}{2b})^{3}=(\frac{2a}{5})^{3}+3(\frac{2a}{5})^{2}(\frac{5}{2b})+3(\frac{2a}{5})(\frac{5}{2b})^{2}+(\frac{5}{2b})^{3} \qquad \Rightarrow\\\\\bold{(\frac{2a}{5}+\frac{5}{2b})^{3}=\frac{8a^{3}}{125}+\frac{6a^{2}}{5b}+\frac{5a}{2b^{2}}+\frac{125}{8b^{3}}}$

$\bold{F)\quad(\frac{2}{5a}+\frac{5}{2b})^{3}}$

$(\frac{2}{5a}+\frac{5}{2b})^{3}=(\frac{2}{5a})^{3}+3(\frac{2}{5a})^{2}(\frac{5}{2b})+3(\frac{2}{5a})(\frac{5}{2b})^{2}+(\frac{5}{2b})^{3} \qquad \Rightarrow\\\\\bold{(\frac{2}{5a}+\frac{5}{2b})^{3}=(\frac{8}{125a^{3}})+\frac{6}{5a^{2}b})+\frac{5}{2ab^{2}}+\frac{125}{8b^{3}}}$