Question

\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }​

Answer 1

Answer:

sinteeta by costeeta! (+) ó86&:jasne-#+@#

Step-by-step explanation:

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Answer 2

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=16$

Explanation:

Given Equation:

  ==>   $\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }$

We have Three values in the equation

==>Let's  Solve it separately

Solve $8^{\frac{1}{3} }$

$\frac{1}{3}$  in the power is the cube root

so, we can write it as,

$8^{\frac{1}{3} }$ = $\sqrt[3]{8}$

$\sqrt[3]{8} = \sqrt[3]{2X2X2}$

$8^{\frac{1}{3} }$ = 2

==> Solve $16^{\frac{1}{3} }$

$16^{\frac{1}{3} }$ = $\sqrt[3]{16}$

$\sqrt[3]{16} = \sqrt[3]{2X2X2X2}$

$16^{\frac{1}{3} }$ = 2√2

==> Solve $32^{\frac{-1}{3} }$

==> $32^{(-1)(\frac{1}{3} )}$

==>  $\sqrt[3]{32}^{-1}$

==> $\sqrt[3]{32}^{-1} = (\sqrt[3]{2X2X2X2})^{-1}$

==>  $\sqrt[3]{32}^{-1} = \sqrt[3]{2X2X2X2X2}^{-1}$

==>  $\sqrt[3]{32}^{-1}=2\sqrt{4}^{-1}$

Substitute all the values

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=\frac{2\times2\sqrt{2} }{2\sqrt{4^{-1} } }$

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=4\sqrt{2} \times2\sqrt{4}$

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=8\sqrt[3]{8}$

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=8\times2$

$\frac{ 8 ^ { ^ { \frac{ 1 }{ 3 } } } \times 16 ^ { \frac{ 1 }{ 3 } } }{ 32 ^ { \frac{ -1 }{ 3 } } }=16$