Question

please help me with this ​

Answer 1

Answer:

Here, The resulting outcome after simplication of the given question is 2744.

Step-by-step explanation:

Here, As per our given question,

=(root260-root64)^3/2 (root260+root64)^3/2

Now here identity which is used:(a+b) (a-b)=a^2-b^2

So,=[(root260)^2-(root64)^2]^3/2

Aftering doing square of roots given here, we get,

=(260-64)^3/2

=(194)^3/2

=(2^2×7^2)^3/2

=(2×7)^3/2×2

=(14)^3 (on multiplication 2 gets cancelled)

=2744

So, Here the correct answer is 2744.

Thank you.

Answer 2

Question -

Solve-

$( \sqrt{260} - \sqrt{64} ) {}^{ \frac{3}{2} } ( \sqrt{260} + \sqrt{64} ) {}^{ \frac{3}{2} }$

Solution -

$= ( \sqrt{260} - \sqrt{64} ) {}^{ \frac{3}{2} } ( \sqrt{260} + \sqrt{64} )^{ \frac{3}{2} }$

$= [( \sqrt{260} - \sqrt{64} )( \sqrt{260} + \sqrt{64} ) ] {}^{ \frac{3}{2} }$

Using the formula (a+b) (a-b) = a²-b²

$= [( { \sqrt{260} )}^{2} - ( { \sqrt{64}) }^{2} ] {}^{ \frac{3}{2} }$

$= (260 - 64) {}^{ \frac{3}{2} }$

$= (196) {}^{ \frac{3}{2} }$

$= {(14)}^{2 \times \frac{3}{2} }$

$= {(14)}^{3}$

$= 2744$

Related Indentities-

$1) {a}^{p} \times { a}^{q} = {a}^{p + q}$

$2) {( {a}^{p} )}^{q} = {a}^{pq}$

$3) \frac{ {a}^{p} }{ {a}^{q} } = {a}^{p - q}$

$4) { a}^{p} {b}^{p} = (ab) {}^{p}$