Answer: \(32-80x+80x^2-40x^3+10x^4-x^5\); \(x=0\) or \(x=2\).

(a) Use the binomial expansion:

\((2-x)^5 =2^5+5(2^4)(-x)+10(2^3)(-x)^2+10(2^2)(-x)^3+5(2)(-x)^4+(-x)^5.\)

So

\((2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5.\)

(b) Combine the powers of \(e\) on the left:

\(\frac{e^{(2-x)^5}\times e^{80x}}{e^{10x^4+32}} =e^{(2-x)^5+80x-(10x^4+32)}.\)

Using the expansion from part (a), the exponent is

\(32-80x+80x^2-40x^3+10x^4-x^5+80x-10x^4-32.\)

This simplifies to

\(80x^2-40x^3-x^5.\)

The equation becomes

\(e^{80x^2-40x^3-x^5}=e^{-x^5}.\)

Therefore

\(80x^2-40x^3-x^5=-x^5.\)

So

\(80x^2-40x^3=0.\)

Factorising,

\(40x^2(2-x)=0.\)

Hence

\(x=0\quad\text{or}\quad x=2.\)