Question

1)(27)^2/3 ÷ (81/16)^1/4
2)(36x^2)^1/2
(27x^-3y^6)^2/6
3) 5√x^20y^10z^5 square root finished ÷x^3/y^3
4)(256a^16/81b^4)^-3/4

Answer 1

$27^{ \frac{2}{3} } / (\frac{81}{16})^{\frac{1}{4}} = [(3)^3]^{\frac{2}{3}} / \frac{ 81^\frac{1}{4} } {{16}^\frac{1}{4}} = (3)^{3*\frac{2}{3} } * 16^{\frac{1}{4}} / 81^{\frac{1}{4}} \\ \\ 3^2 * (2^4)^{\frac{1}{4}} / [(3)^{4}]^{\frac{1}{4}} = 3^2 * 2^{4*\frac{1}{4}} / 3^{4*\frac{1}{4}} = 3^2*2^1 / 3^1 = 6$

2)

$(36 * x^2)^\frac{1}{2} (27 * x^{-3} * y^6)^{\frac{2}{6}} = (6^2)^{\frac{1}{2}} * (x^2)^{\frac{1}{2}} [3^3 * x^{-3} *y^6)]^{\frac{1}{3}} \\ \\ = 6^{2*\frac{1}{2}} * x^{2 *{\frac{1}{2}}} * [3^{3*\frac{1}{3}} * x^{-3*\frac{1}{3}} * y^{6 * \frac{1}{3}}] \\ \\ = 6^1 *x^1 * 3^1 * x^{-1} * y^2 = 6 * 3 * x^{1-1} * y^2 = 18*x^0 * y^2 \\ \\ = 18 y^2$

3)

$5 * \sqrt{x^{20}*y^{10}*z^5} / \frac{x^3}{y^3} =$

$5 * x^{20*\frac{1}{2}} * y^{10*\frac{1}{2}}* z^{5*\frac{1}{2}} * \frac{y^3}{x^3} = 5*x^{10}*y^{5}*z^{\frac{5}{2}} * y^3 / x^3 \\ \\ 5*x^{10-3}*y^{5+3}*z^{2+\frac{1}{2}} \\ \\ 5 x^7* y^8 * z^2*z^{\frac{1}{2}} = 5x^7y^8z^2\sqrt{z}$

4 )

$(256 a^{16} / 81 b^4)^{-\frac{3}{4}} = \frac{1}{(256 a^{16} / 81 b^4)^{-\frac{3}{4}}} \\ \\ = (81b^4 / 256 a^{16})^{+\frac{3}{4}} = \frac{(81b^4)^\frac{3}{4}}{(256a^{16})^{\frac{3}{4}}} \\ \\ = \frac{3^{4*\frac{3}{4}}*b^{4*\frac{3}{4}}}{4^{4*\frac{3}{4}} *a^{16*\frac{3}{4}}} \\ \\ = \frac{3^3*b^3}{4^3*a^{12}}$