Question

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

Answer:

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Step-by-step explanation:

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Answer 2

The answer of this question is 16.

Step-by-step explanation:

We are given,

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

and we have to solve this and get the answer of this question

• Formula used,

1. $x^{a} *x^{b}=x^{a+b}$
2. $\frac{x^{a}}{x^{b}} = x^{a-b}$

• Solving given term,

we have,

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

we can write 8, 16 and 32 as

$8=2*2*2$

$16=2*2*2*2$

$32=2*2*2*2*2$

or $8=2^{3}$

$16=2^{4}$

$32=2^{5}$

by writing this we can make bases of power same.

new equation we get is,

$\frac{(2^{3}) ^{\frac{1}{3} } *(2^{4} )^{\frac{1}{3} }}{(2^{5} )^{\frac{-1}{3} }}$

$\frac{2 ^{\frac{3}{3} } *2^{\frac{4}{3} }}{2^{\frac{-5}{3} }}$

now by using formula(1) our numerator is,

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

            $=2^{\frac{3+4}{3} }$

            $=2^{\frac{7}{3} }$

we get numerator as $2^{\frac{7}{3} }$.

by using new numerator we get the fraction as,

$\frac{2^{\frac{7}{3} }}{2^{\frac{-5}{3} }}$

now by using formula(2),

$2^{\frac{7}{3}-\frac{(-5)}{3} }$

$2^{\frac{7}{3}+\frac{5}{3} }$

$=2^{\frac{7+5}{3} }$

$=2^{\frac{12}{3}}$

$=2^4}$

and $2^{4} =16$

so the answer of this question is 16.