simplify: (((e 10x - e -10x )/2)-((e 10x + e -10x )/2))) 10

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Let  A = e^10x  and let  B = e^-10x.

Ignoring the outside exponent for a while,  the problem becomes [ (A - B) / 2 ] - [ (A + B) / 2 ]

   =  [A/2 - B/2] - [A/2 + B/2]  =  A/2 - B/2 - A/2 - B/2  =  - 2B/2  =  - B     which is     -e^-10x

Bringing the exponent back into the problem:  (-e^-10x)¹⁰  =  e^-100x

because a negative number to an even exponent is positive, and you need to multiply the exponents.

simplify: [ ((e^10x - e^-10x)/2) - ((e^10x + e^-10x)/2) ]¹⁰

need help!!!

\(\begin{array}{rcll} \left( \frac{e^{10x} - e^{-10x} } {2} - \frac{e^{10x} + e^{-10x} } {2} \right)^{10} &=& \left[ \frac{e^{10x} - e^{-10x} -( e^{10x} + e^{-10x} )} {2} \right]^{10} \\\\ &=& \left( \frac{e^{10x} - e^{-10x} - e^{10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{e^{10x}- e^{10x} - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{ 0 - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{ - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{- 2\cdot e^{-10x} } {2} \right)^{10} \\\\ &=& \left( -e^{-10x} \right)^{10} \\\\ &=& \left[ (-1)\cdot e^{-10x} \right]^{10} \\\\ &=& (-1)^{10}\cdot e^{-10\cdot 10 x} \\\\ &=& 1\cdot e^{-10\cdot 10 x} \\\\ &=& 1\cdot e^{-100 x} \\\\ \mathbf{ \left( \frac{e^{10x} - e^{-10x} } {2} - \frac{e^{10x} + e^{-10x} } {2} \right)^{10} } &\mathbf{=}& \mathbf{ e^{-100 x} } \end{array}\)

heureka is a mathmatical boss! Thanks!