Precalculus (6th Edition) Blitzer

The required solution is $5{{x}^{4}}+10{{x}^{3}}h+10{{x}^{2}}{{h}^{2}}+5x{{h}^{3}}+{{h}^{4}}$.
Let us consider the provided ratio: $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Now, put the provided function in the above ratio: \begin{align} & \frac{\left( {{\left( x+h \right)}^{5}}+8 \right)-\left( {{x}^{5}}+8 \right)}{h}=\frac{{{\left( x+h \right)}^{5}}+8-{{x}^{5}}-8}{h} \\ & =\frac{{{\left( x+h \right)}^{5}}-{{x}^{5}}}{h} \\ & =\frac{{{x}^{5}}+5{{x}^{4}}h+10{{x}^{3}}{{h}^{2}}+10{{x}^{2}}{{h}^{3}}+5x{{h}^{4}}+{{h}^{5}}-{{x}^{4}}}{h} \\ & =\frac{h\left( 5{{x}^{4}}+10{{x}^{3}}h+10{{x}^{2}}{{h}^{2}}+5x{{h}^{3}}+{{h}^{4}} \right)}{h} \end{align} So, $\frac{\left( {{\left( x+h \right)}^{5}}+8 \right)-\left( {{x}^{5}}+8 \right)}{h}=5{{x}^{4}}+10{{x}^{3}}h+10{{x}^{2}}{{h}^{2}}+5x{{h}^{3}}+{{h}^{4}}$ Thus, the ratio is $5{{x}^{4}}+10{{x}^{3}}h+10{{x}^{2}}{{h}^{2}}+5x{{h}^{3}}+{{h}^{4}}$. So, it is calculated by substituting the function in the ratio and the value obtained matches with the binomial expansion formula.