Simplify (4j)^3(-j^-2k^2)(3j^5k^-7)^3

${(4 j)}^{3} (- j^{- 2} k^{2}) {(3 j^{5} k^{- 7})}^{3}$

Rewrite using the commutative property of multiplication.

$- {(4 j)}^{3} j^{- 2} k^{2} {(3 j^{5} k^{- 7})}^{3}$

Apply the product rule to $4 j$ .

$- (4^{3} j^{3}) j^{- 2} k^{2} {(3 j^{5} k^{- 7})}^{3}$

Multiply $j^{3}$ by $j^{- 2}$ by adding the exponents.

Move $j^{- 2}$ .

$- (4^{3} (j^{- 2} j^{3})) k^{2} {(3 j^{5} k^{- 7})}^{3}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- (4^{3} j^{- 2 + 3}) k^{2} {(3 j^{5} k^{- 7})}^{3}$

Add $- 2$ and $3$ .

$- (4^{3} j^{1}) k^{2} {(3 j^{5} k^{- 7})}^{3}$

Simplify $- (4^{3} j^{1}) k^{2} {(3 j^{5} k^{- 7})}^{3}$ .

$- (4^{3} j) k^{2} {(3 j^{5} k^{- 7})}^{3}$

Raise $4$ to the power of $3$ .

$- (64 j) k^{2} {(3 j^{5} k^{- 7})}^{3}$

Multiply $64$ by $- 1$ .

$- 64 j k^{2} {(3 j^{5} k^{- 7})}^{3}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 64 j k^{2} {(3 j^{5} \frac{1}{k^{7}})}^{3}$

Multiply $3 j^{5} \frac{1}{k^{7}}$ .

Combine $3$ and $\frac{1}{k^{7}}$ .

$- 64 j k^{2} {(j^{5} \frac{3}{k^{7}})}^{3}$

Combine $j^{5}$ and $\frac{3}{k^{7}}$ .

$- 64 j k^{2} {(\frac{j^{5} \cdot 3}{k^{7}})}^{3}$

Move $3$ to the left of $j^{5}$ .

$- 64 j k^{2} {(\frac{3 \cdot j^{5}}{k^{7}})}^{3}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $\frac{3 j^{5}}{k^{7}}$ .

$- 64 j k^{2} \frac{{(3 j^{5})}^{3}}{{(k^{7})}^{3}}$

Apply the product rule to $3 j^{5}$ .

$- 64 j k^{2} \frac{3^{3} {(j^{5})}^{3}}{{(k^{7})}^{3}}$

Simplify the numerator.

Raise $3$ to the power of $3$ .

$- 64 j k^{2} \frac{27 {(j^{5})}^{3}}{{(k^{7})}^{3}}$

Multiply the exponents in ${(j^{5})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 64 j k^{2} \frac{27 j^{5 \cdot 3}}{{(k^{7})}^{3}}$

Multiply $5$ by $3$ .

$- 64 j k^{2} \frac{27 j^{15}}{{(k^{7})}^{3}}$

Simplify terms.

Multiply the exponents in ${(k^{7})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 64 j k^{2} \frac{27 j^{15}}{k^{7 \cdot 3}}$

Multiply $7$ by $3$ .

$- 64 j k^{2} \frac{27 j^{15}}{k^{21}}$

Cancel the common factor of $k^{2}$ .

Factor $k^{2}$ out of $- 64 j k^{2}$ .

$k^{2} (- 64 j) \frac{27 j^{15}}{k^{21}}$

Factor $k^{2}$ out of $k^{21}$ .

$k^{2} (- 64 j) \frac{27 j^{15}}{k^{2} k^{19}}$

Cancel the common factor.

$k^{2} (- 64 j) \frac{27 j^{15}}{k^{2} k^{19}}$

Rewrite the expression.

$- 64 j \frac{27 j^{15}}{k^{19}}$

Combine $- 64$ and $\frac{27 j^{15}}{k^{19}}$ .

$j \frac{- 64 (27 j^{15})}{k^{19}}$

Multiply $27$ by $- 64$ .

$j \frac{- 1728 j^{15}}{k^{19}}$

Combine $j$ and $\frac{- 1728 j^{15}}{k^{19}}$ .

$\frac{j (- 1728 j^{15})}{k^{19}}$

Raise $j$ to the power of $1$ .

$\frac{- 1728 (j^{1} j^{15})}{k^{19}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 1728 j^{1 + 15}}{k^{19}}$

Add $1$ and $15$ .

$\frac{- 1728 j^{16}}{k^{19}}$

Move the negative in front of the fraction.

$- \frac{1728 j^{16}}{k^{19}}$

Simplify (4j)^3(-j^-2k^2)(3j^5k^-7)^3

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