Question

express the following fraction in simplest form, only using positive exponents.\\(\frac{(-2hk^4)^3}{2h^{-8}k^{-8}}\\)

Explanation:

Step1: Expand the numerator

Using the power of a product rule \((ab)^n = a^n b^n\) and \((a^m)^n=a^{mn}\), we have \((-2hk^4)^3=(-2)^3h^3(k^4)^3=-8h^3k^{12}\).

Step2: Simplify the fraction

Now the expression is \(\frac{-8h^3k^{12}}{2h^{-8}k^{-8}}\). Divide the coefficients: \(\frac{-8}{2}=-4\). For the \(h\) terms, use the quotient rule \(a^m\div a^n = a^{m - n}\), so \(h^3\div h^{-8}=h^{3-(-8)} = h^{11}\). For the \(k\) terms, \(k^{12}\div k^{-8}=k^{12-(-8)}=k^{20}\).

Step3: Combine the results

Multiply the coefficient and the variable terms together: \(-4h^{11}k^{20}\).

Answer:

\(-4h^{11}k^{20}\)