Resolver (j)/(-7)^3+(-2)^3-[(-21)+35-sqrt[3]{125}*(-8)]

$$( j ) / ( - 7 ) ^ { 3 } + ( - 2 ) ^ { 3 } - [ ( - 21 ) + 35 - \sqrt[ 3 ] { 125 } \times ( - 8 ) ]$$

Fatorizar

$$\frac{-j-21266}{343}$$

Passos Para a Resolução

Calcule -7 elevado a 3 e obtenha -343.

$$factor(\frac{j}{-343}+\left(-2\right)^{3}-\left(-21+35-\sqrt[3]{125}\left(-8\right)\right))$$

Calcule -2 elevado a 3 e obtenha -8.

$$factor(\frac{j}{-343}-8-\left(-21+35-\sqrt[3]{125}\left(-8\right)\right))$$

Some -21 e 35 para obter 14.

$$factor(\frac{j}{-343}-8-\left(14-\sqrt[3]{125}\left(-8\right)\right))$$

Calcule \sqrt[3]{125} e obtenha 5.

$$factor(\frac{j}{-343}-8-\left(14-5\left(-8\right)\right))$$

Multiplique 5 e -8 para obter -40.

$$factor(\frac{j}{-343}-8-\left(14-\left(-40\right)\right))$$

O oposto de -40 é 40.

$$factor(\frac{j}{-343}-8-\left(14+40\right))$$

Some 14 e 40 para obter 54.

$$factor(\frac{j}{-343}-8-54)$$

Subtraia 54 de -8 para obter -62.

$$factor(\frac{j}{-343}-62)$$

Decomponha \frac{1}{343}.

$$\frac{-j-21266}{343}$$

Avaliar

$$-\frac{j}{343}-62$$