Resolver {[(1/2a-2/3b)(1/2a+2/3b)^{3}-(frac{1}{4}a^{2}-frac{4}{9}b^{2})(frac{4}{9}b^{2}+frac{1}{4}a^2)]-frac{1}{3}ab(frac{1}{2}a^2+frac{1}{9}b^2)}^3

$$\{ [ ( \frac { 1 } { 2 } a - \frac { 2 } { 3 } b ) ( \frac { 1 } { 2 } a + \frac { 2 } { 3 } b ) ^ { 3 } - ( \frac { 1 } { 4 } a ^ { 2 } - \frac { 4 } { 9 } b ^ { 2 } ) ( \frac { 4 } { 9 } b ^ { 2 } + \frac { 1 } { 4 } a ^ { 2 } ) ] - \frac { 1 } { 3 } a b ( \frac { 1 } { 2 } a ^ { 2 } + \frac { 1 } { 9 } b ^ { 2 } ) \} ^ { 3 }$$

Calcular

$$-\frac{a^{3}b^{9}}{27}$$

Pasos da solución

Usar teorema binomial \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} para expandir \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3}.

$$\left(\left(\frac{1}{2}a-\frac{2}{3}b\right)\left(\frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3}\right)-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Usa a propiedade distributiva para multiplicar \frac{1}{2}a-\frac{2}{3}b por \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} e combina os termos semellantes.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Considera \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). A multiplicación pódese transformar na diferencia de cadrados mediante a regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Expande \left(\frac{1}{4}a^{2}\right)^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}\left(a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 2 e 2 para obter 4.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Calcula \frac{1}{4} á potencia de 2 e obtén \frac{1}{16}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Expande \left(\frac{4}{9}b^{2}\right)^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}\left(b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 2 e 2 para obter 4.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Calcula \frac{4}{9} á potencia de 2 e obtén \frac{16}{81}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\frac{16}{81}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para calcular o oposto de \frac{1}{16}a^{4}-\frac{16}{81}b^{4}, calcula o oposto de cada termo.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\frac{1}{16}a^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Combina \frac{1}{16}a^{4} e -\frac{1}{16}a^{4} para obter 0.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Combina -\frac{16}{81}b^{4} e \frac{16}{81}b^{4} para obter 0.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Usa a propiedade distributiva para multiplicar -\frac{1}{3}ab por \frac{1}{2}a^{2}+\frac{1}{9}b^{2}.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{6}a^{3}b-\frac{1}{27}ab^{3}\right)^{3}$$

Combina \frac{1}{6}a^{3}b e -\frac{1}{6}a^{3}b para obter 0.

$$\left(-\frac{8}{27}ab^{3}-\frac{1}{27}ab^{3}\right)^{3}$$

Combina -\frac{8}{27}ab^{3} e -\frac{1}{27}ab^{3} para obter -\frac{1}{3}ab^{3}.

$$\left(-\frac{1}{3}ab^{3}\right)^{3}$$

Expande \left(-\frac{1}{3}ab^{3}\right)^{3}.

$$\left(-\frac{1}{3}\right)^{3}a^{3}\left(b^{3}\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 3 e 3 para obter 9.

$$\left(-\frac{1}{3}\right)^{3}a^{3}b^{9}$$

Calcula -\frac{1}{3} á potencia de 3 e obtén -\frac{1}{27}.

$$-\frac{1}{27}a^{3}b^{9}$$

Expandir

$$-\frac{a^{3}b^{9}}{27}$$

Pasos da solución

Usar teorema binomial \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} para expandir \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3}.

$$\left(\left(\frac{1}{2}a-\frac{2}{3}b\right)\left(\frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3}\right)-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Usa a propiedade distributiva para multiplicar \frac{1}{2}a-\frac{2}{3}b por \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} e combina os termos semellantes.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Considera \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). A multiplicación pódese transformar na diferencia de cadrados mediante a regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Expande \left(\frac{1}{4}a^{2}\right)^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}\left(a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 2 e 2 para obter 4.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Calcula \frac{1}{4} á potencia de 2 e obtén \frac{1}{16}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Expande \left(\frac{4}{9}b^{2}\right)^{2}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}\left(b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 2 e 2 para obter 4.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Calcula \frac{4}{9} á potencia de 2 e obtén \frac{16}{81}.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\frac{16}{81}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Para calcular o oposto de \frac{1}{16}a^{4}-\frac{16}{81}b^{4}, calcula o oposto de cada termo.

$$\left(\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\frac{1}{16}a^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Combina \frac{1}{16}a^{4} e -\frac{1}{16}a^{4} para obter 0.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Combina -\frac{16}{81}b^{4} e \frac{16}{81}b^{4} para obter 0.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)\right)^{3}$$

Usa a propiedade distributiva para multiplicar -\frac{1}{3}ab por \frac{1}{2}a^{2}+\frac{1}{9}b^{2}.

$$\left(\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{6}a^{3}b-\frac{1}{27}ab^{3}\right)^{3}$$

Combina \frac{1}{6}a^{3}b e -\frac{1}{6}a^{3}b para obter 0.

$$\left(-\frac{8}{27}ab^{3}-\frac{1}{27}ab^{3}\right)^{3}$$

Combina -\frac{8}{27}ab^{3} e -\frac{1}{27}ab^{3} para obter -\frac{1}{3}ab^{3}.

$$\left(-\frac{1}{3}ab^{3}\right)^{3}$$

Expande \left(-\frac{1}{3}ab^{3}\right)^{3}.

$$\left(-\frac{1}{3}\right)^{3}a^{3}\left(b^{3}\right)^{3}$$

Para elevar unha potencia a outra potencia, multiplica os expoñentes. Multiplica 3 e 3 para obter 9.

$$\left(-\frac{1}{3}\right)^{3}a^{3}b^{9}$$

Calcula -\frac{1}{3} á potencia de 3 e obtén -\frac{1}{27}.

$$-\frac{1}{27}a^{3}b^{9}$$