Question

(✓32-√5)^1/3(√32+√5)^1/3​

Answer 1

Answer:

\red { ( \sqrt{32} - \sqrt{5} )^{\frac{1}{3}} \times ( \sqrt{32} + \sqrt{5} )^{\frac{1}{3}} }

=  \Big(( \sqrt{32} - \sqrt{5} )( \sqrt{32} + \sqrt{5} )\Big)^{\frac{1}{3}}  

\blue { We\:are \:using \: following \: formulae :}

 \pink {i)a^{m} \times b^{m} = (ab)^{m} }

\pink {ii) (a^{m})^{n} = a^{m \times n }}\\\pink { iii) (a + b )(a - b) = a^{2} - b^{2} }

= [ (\sqrt{32})^{2} - (\sqrt{5})^{2})^{\frac{1}{3}}

= ( 32 - 5 )^{\frac{1}{3}} \\= 27^{\frac{1}{3}}\\= (3^{3})^{\frac{1}{3}} \\= (3)^{3 \times \frac{1}{3}} \\= 3

Answer 2

Answer:

Given: The term (√32-√5)^1/3 x (√32+√5)^1/3

To find: The product of the above term

Solution:

So we have provided with the term:

             (√32-√5)^1/3 x (√32+√5)^1/3

Now by using the formula: a^m x b^m = (ab)^m

So applying this formula, we get:

             { (√32-√5) x (√32+√5) }^1/3

Now by using the formula: (a - b)(a + b) = a^2 - b^2

So applying this formula, we get:

            { (√32)²-(√5)² }^1/3

Now solving this we get:

            { 32 - 5 }^1/3

            { 27 }^1/3

Now cubic root of 27 is 3, so:

            { 3^3 }^1/3  = 3

Answer:

          So the product of (√32-√5)^1/3 x (√32+√5)^1/3 is 3.

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