Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Superscript 5 )squared (8 n Superscript 4 Baseline q )Superscript negative 2 Over (negative 3 nq Superscript 9 )Superscript negative 1 Baseline (4 n cubed q Superscript 9 )cubed EndFraction StartFraction

Answer:[tex] \frac{ - 3}{ 256 {q}^{10} {n}^{8} } [/tex]Step by step explanation:[tex] \frac{ {(4 {n}^{4} {q}^{5})}^{2} {(8 {n}^{4} q)}^{-2} }{ {(- 3 {nq}^{9})}^{ - 1} {(4 {n}^{3} {q}^{9}) }^{3} } [/tex]first we will change the terms with negative superscrips to the other side of the fraction[tex] \frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} } [/tex]then we will distribute the superscripts[tex] \frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2} {q}^{2} } [/tex][tex] \frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8} {q}^{2} } [/tex]as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents[tex] {4}^{2 - 3} {q}^{10 + 9 - 2 - 27} {n}^{8 + 1 - 8 - 9} {8}^{ - 2} { (- 3)}^{1} [/tex][tex] {4}^{ - 1} {q}^{ - 10} {n}^{ - 8} {8}^{ - 2} { (- 3)}^{1} [/tex]then we will change again the terms with negative superscrips to the other side of the fraction[tex] \frac{ - 3}{ 4 \times {8}^{2} {q}^{10} {n}^{8} } [/tex][tex] \frac{ - 3}{ 256 {q}^{10} {n}^{8} } [/tex]