Answer
$$\left( {\bf{a}} \right)\frac{{3{x^{5/3}}}}{5} + C,\left( {\bf{b}} \right) - \frac{1}{{{x^5}}} + C,\left( {\bf{c}} \right)8{x^{1/8}} + C$$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)\int {\root 3 \of {{x^2}} } dx \cr
& {\text{rewriting the integrand}}{\text{,}}\,\,{\text{use }}\root n \of {{x^m}} = {x^{m/n}} \cr
& \,\,\, = \int {{x^{2/3}}} dx \cr
& {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{2/3}}} dx = \frac{{{x^{2/3 + 1}}}}{{2/3 + 1}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{2/3}}} dx = \frac{{{x^{5/3}}}}{{5/3}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{2/3}}} dx = \frac{{3{x^{5/3}}}}{5} + C \cr
& \cr
& \left( {\bf{b}} \right)\int {\frac{1}{{{x^6}}}} dx \cr
& {\text{rewriting the integrand}}{\text{,}}\,\,{\text{use }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& \,\,\,\,\,\, = \int {{x^{ - 6}}} dx \cr
& {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \,\,\,\,\,\,\,\int {{x^{ - 6}}} dx = \frac{{{x^{ - 6 + 1}}}}{{ - 6 + 1}} + C \cr
& \,\,\,\,\,\,\,\int {{x^{ - 6}}} dx = \frac{{{x^{ - 5}}}}{{ - 5}} + C \cr
& {\text{rewriting}} \cr
& \,\,\,\,\,\,\,\int {{x^{ - 6}}} dx = - \frac{1}{{{x^5}}} + C \cr
& \cr
& \left( {\bf{c}} \right)\int {{x^{ - 7/8}}} dx \cr
& {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{ - 7/8}}} dx = \frac{{{x^{ - 7/8 + 1}}}}{{ - 7/8 + 1}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{ - 7/8}}} dx = \frac{{{x^{1/8}}}}{{1/8}} + C \cr
& \,\,\,\,\,\,\,\,\int {{x^{ - 7/8}}} dx = 8{x^{1/8}} + C \cr} $$
