Answer:
(i) \[\frac{{{2}^{3}}\times {{3}^{4}}\times 4}{3\times 32}=\frac{{{2}^{3}}\times {{3}^{4}}\times {{2}^{2}}}{3\times {{2}^{5}}}=\frac{{{2}^{3+2}}\times {{3}^{4}}}{{{2}^{5}}\times 3}\] \[=\frac{{{2}^{5}}\times {{3}^{4}}}{{{2}^{5}}\times 3}=\frac{{{3}^{4}}}{{{3}^{1}}}={{3}^{4-1}}={{3}^{3}}\] (ii) \[\{{{({{5}^{2}})}^{3}}\times {{5}^{4}}\}\div {{5}^{7}}=({{5}^{2\times 3}}\times {{5}^{4}})\div {{5}^{7}}\] \[=({{5}^{6}}\times {{5}^{4}})\div {{5}^{7}}={{5}^{6+4}}\div {{5}^{7}}\] \[={{5}^{10}}\div {{5}^{7}}={{5}^{10-7}}={{5}^{3}}\] (iii) \[{{25}^{4}}\div {{5}^{3}}={{({{5}^{2}})}^{4}}\div {{5}^{3}}={{5}^{2\times 4}}\div {{5}^{3}}\] \[={{5}^{8}}\div {{5}^{3}}={{5}^{8-3}}={{5}^{5}}\] (iv) \[\frac{3\times {{7}^{2}}\times {{11}^{8}}}{21\times {{11}^{3}}}=\frac{3\times {{7}^{2}}\times {{11}^{8}}}{3\times 7\times {{11}^{3}}}\] \[=\frac{{{7}^{2}}\times {{11}^{8}}}{7\times {{11}^{3}}}={{7}^{2-1}}\times {{11}^{8-3}}={{7}^{1}}\times {{11}^{5}}\] (v) \[\frac{{{3}^{7}}}{{{3}^{4}}\times {{3}^{3}}}=\frac{{{3}^{7}}}{{{3}^{4+3}}}=\frac{{{3}^{7}}}{{{3}^{7}}}={{3}^{7-7}}={{3}^{0}}=1\] (vi) \[{{2}^{0}}+{{3}^{0}}+{{4}^{0}}=1+1+1=3\] (vii) \[{{2}^{0}}\times {{3}^{0}}\times {{4}^{0}}=1\times 1\times 1=1\] (viii) \[({{3}^{0}}+{{2}^{0}})\times {{5}^{0}}=(1+1)\times 1=2\times 1=2\] (ix) \[\frac{{{2}^{8}}\times {{a}^{5}}}{{{4}^{3}}\times {{a}^{3}}}=\frac{{{2}^{8}}\times {{a}^{5}}}{{{({{2}^{2}})}^{3}}\times {{a}^{3}}}=\frac{{{2}^{8}}\times {{a}^{5}}}{{{2}^{2\times 3}}\times {{a}^{3}}}\] \[=\frac{{{2}^{8}}\times {{a}^{5}}}{{{2}^{6}}\times {{a}^{3}}}={{2}^{8-6}}\times {{a}^{5-3}}={{2}^{2}}\times {{a}^{2}}\] (x) \[\left( \frac{{{a}^{5}}}{{{a}^{3}}} \right)\times {{a}^{8}}={{a}^{5-3}}\times {{a}^{8}}={{a}^{2}}\times {{a}^{8}}={{a}^{2+8}}={{a}^{10}}\] (xi) \[\frac{{{4}^{5}}\times {{a}^{8}}{{b}^{3}}}{{{4}^{5}}\times {{a}^{5}}{{b}^{2}}}=\frac{{{a}^{8}}{{b}^{3}}}{{{a}^{5}}{{b}^{2}}}={{a}^{8-5}}{{b}^{3-2}}={{a}^{3}}{{b}^{1}}\] (xii) \[{{({{2}^{3}}\times 2)}^{2}}={{({{2}^{3}}\times {{2}^{1}})}^{2}}\] \[={{({{2}^{3+1}})}^{2}}={{({{2}^{4}})}^{2}}={{2}^{4\times 2}}={{2}^{8}}\]
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