Exponents problem #2

$\sqrt {8} \times 8^\frac {1}{4} \times \left (2^\frac {1}{4} \right)^3 - \sqrt {32} \times \sqrt {2}$

$= \sqrt {2^3} \times 2^\frac {3}{4} \times 2^\frac {3}{4} - \sqrt {64}$

$= 2^\frac {3}{2} \times 2^\frac {3}{4} \times 2^\frac {3}{4} - 8$

$= 2^{\frac {3}{2} + \frac {3}{4} + \frac {3}{4}} - 2^3$

$= 2^3 - 2^3$

$= 0$

$\sqrt{8} \times 8^\frac{1}{4} \times \left(2^\frac{1}{4}\right)^3 - \sqrt{32} \times \sqrt{2}$

$= 8^\frac{2}{4} \times 8^\frac{1}{4} \times 8^\frac{1}{4} - \sqrt{64}$

$= 8 - 8$

$= 0$

$\sqrt {8}×8^{\frac{1}{4}}×(2^{\frac{1}{4}})^3-\sqrt {32}×\sqrt 2\\=2^{\frac{3}{2}}×2^{\frac{3}{4}}×2^{\frac{3}{4}}-2^{\frac{5}{2}}×2^{\frac{1}{2}}\\=(2^3)-(2^3)=0$

=(2)^3/2 X (2)^3/4 X(2)^3/4 X -(2)^5/2 X (2)^1/2 =(2)^3/2+3/4+3/4 - (2)^5/2+1/2 =(2)^6+3+3/4 - (2)^5+1/2 =(2)^3 - (2)^3 =0