Question

Find the error & explain why it is wrong:

Answer 1

$( { \frac{4 {x}^{3} }{ {y}^{4} }) }^{ - 2} = \frac{16x {}^{6} }{ {y}^{8} }$

$\frac{(4 {x}^{3})^{ - 2} }{ { {(y}^{4} )}^{ - 2} } = \frac{16 {x}^{6} }{ {y}^{8} }$

$\frac{ {16x}^{ - 6} }{ {y }^{ - 8} } = \frac{ {16x}^{6} }{ {y}^{8} }$

The power of -2 was mistook as 2.