Rešite {[(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

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

Ovrednoti

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

Koraki rešitve

Uporabite binomski izrek \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3}, da razširite \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3}.

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

Uporabite lastnost distributivnosti za množenje \frac{1}{2}a-\frac{2}{3}b krat \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} in kombiniranje pogojev podobnosti.

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

Razmislite o \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). Množenje je lahko preoblikovano v razliko kvadratov s pravilom: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 2 in 2, da dobite 4.

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

Izračunajte potenco \frac{1}{4} števila 2, da dobite \frac{1}{16}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 2 in 2, da dobite 4.

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

Izračunajte potenco \frac{4}{9} števila 2, da dobite \frac{16}{81}.

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

Če želite poiskati nasprotno vrednost za \frac{1}{16}a^{4}-\frac{16}{81}b^{4}, poiščite nasprotno vrednost vsakega izraza.

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

Združite \frac{1}{16}a^{4} in -\frac{1}{16}a^{4}, da dobite 0.

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

Združite -\frac{16}{81}b^{4} in \frac{16}{81}b^{4}, da dobite 0.

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

Uporabite distributivnost, da pomnožite -\frac{1}{3}ab s/z \frac{1}{2}a^{2}+\frac{1}{9}b^{2}.

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

Združite \frac{1}{6}a^{3}b in -\frac{1}{6}a^{3}b, da dobite 0.

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

Združite -\frac{8}{27}ab^{3} in -\frac{1}{27}ab^{3}, da dobite -\frac{1}{3}ab^{3}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 3 in 3, da dobite 9.

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

Izračunajte potenco -\frac{1}{3} števila 3, da dobite -\frac{1}{27}.

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

Razširi

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

Koraki rešitve

Uporabite binomski izrek \left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3}, da razširite \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3}.

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

Uporabite lastnost distributivnosti za množenje \frac{1}{2}a-\frac{2}{3}b krat \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} in kombiniranje pogojev podobnosti.

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

Razmislite o \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). Množenje je lahko preoblikovano v razliko kvadratov s pravilom: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 2 in 2, da dobite 4.

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

Izračunajte potenco \frac{1}{4} števila 2, da dobite \frac{1}{16}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 2 in 2, da dobite 4.

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

Izračunajte potenco \frac{4}{9} števila 2, da dobite \frac{16}{81}.

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

Če želite poiskati nasprotno vrednost za \frac{1}{16}a^{4}-\frac{16}{81}b^{4}, poiščite nasprotno vrednost vsakega izraza.

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

Združite \frac{1}{16}a^{4} in -\frac{1}{16}a^{4}, da dobite 0.

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

Združite -\frac{16}{81}b^{4} in \frac{16}{81}b^{4}, da dobite 0.

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

Uporabite distributivnost, da pomnožite -\frac{1}{3}ab s/z \frac{1}{2}a^{2}+\frac{1}{9}b^{2}.

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

Združite \frac{1}{6}a^{3}b in -\frac{1}{6}a^{3}b, da dobite 0.

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

Združite -\frac{8}{27}ab^{3} in -\frac{1}{27}ab^{3}, da dobite -\frac{1}{3}ab^{3}.

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

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

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

Če želite potenco potencirati z drugo potenco, pomnožite eksponente. Pomnožite 3 in 3, da dobite 9.

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

Izračunajte potenco -\frac{1}{3} števila 3, da dobite -\frac{1}{27}.

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