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Answer 1 · 2018-03-31T08:49:39

E

Explanation:

$\frac{b^{3} \sqrt[3]{a^{2} b^{5}}}{a}$ $(b^3root(3)(a^2b^5))/a$ this is what your question looks like as

Rule 1: $a^{- 1} = \frac{1}{a^{1}} = \frac{1}{a}$ $a^-1=1/a^1=1/a$

Rule 2: $\sqrt{x} = x^{\frac{1}{2}}$ $sqrtx=x^(1/2)$

$\frac{b^{2} {(a^{2} b^{5})}^{\frac{1}{3}}}{a}$ $(b^2(a^2b^5)^(1/3))/a$

Rule 3: $\sqrt{a b} = \sqrt{a} \sqrt{b} = {(a b)}^{\frac{1}{2}} = a^{\frac{1}{2}} b^{\frac{1}{2}}$ $sqrt(ab)=sqrtasqrtb=(ab)^(1/2)=a^(1/2)b^(1/2)$

$\frac{b^{2} a^{\frac{2}{3}} b^{\frac{5}{3}}}{a}$ $(b^2a^(2/3)b^(5/3))/a$

Rule 4: $a^{2} \cdot a^{3} = a^{2 + 3} = a^{5}$ $a^2*a^3=a^(2+3)=a^5$

Rule 5: $\frac{a^{2}}{a^{3}} = a^{2 - 3} = a^{- 1}$ $a^2/a^3=a^(2-3)=a^-1$

$b^{2 + \frac{5}{3}} a^{\frac{2}{3} - 1} = b^{\frac{6}{3} + \frac{5}{3}} a^{\frac{2}{3} - \frac{3}{3}} = b^{\frac{11}{3}} a^{- \frac{1}{3}} = \frac{b^{\frac{11}{3}}}{a^{\frac{1}{3}}}$ $b^(2+5/3)a^(2/3-1)=b^(6/3+5/3)a^(2/3-3/3)=b^(11/3)a^(-1/3)=b^(11/3)/a^(1/3)$

So the answer is E

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