Question

The value of \(\huge \frac { { (4+\sqrt { 15 } ) }^{ \frac { 3 }{ 2 } }+{ (4-\sqrt { 15 } ) }^{ \frac { 3 }{ 2 } } }{ { (6+\sqrt { 35 } ) }^{ \frac { 3 }{ 2 } }-{ (6-\sqrt { 35 } ) }^{ \frac { 3 }{ 2 } } } \) can be expressed in the form \(\Large \frac{a}{b}\), when a and b are positive, coprime integers. Determine the value of \(\large a+b\).

Answer 1

@ganeshie8 Help plzz!!

Answer 2

@vishweshshrimali5

Answer 3

Do you have any ideas?

Answer 4

are there any way to factorize?

Answer 5

should be "is"

Answer 6

whats the name of the unit(lesson) this question was in?

Answer 7

yes thats a good idea ! use below to factor numerator and denominator
\[\large a^3 + b^3 = (a+b)(a^2+b^2-ab)\]
\[\large a^3 - b^3 = (a-b)(a^2+b^2+ab)\]

Answer 8

\[\large \dfrac{ \left(\sqrt{4+\sqrt{15}} +\sqrt{4-\sqrt{15}}\right)\left(8-1\right) }{ \left(\sqrt{6+\sqrt{35}} -\sqrt{6-\sqrt{35}}\right)\left(12+1\right) }\]

Answer 9

\[\large \dfrac{ \left(\frac{1}{2}(\sqrt{5} + \sqrt{3}) + \frac{1}{2}(\sqrt{5} - \sqrt{3})\right)7 }{ \left(\frac{1}{2}(\sqrt{7} + \sqrt{5}) - \frac{1}{2}(\sqrt{7} - \sqrt{5})\right)13 }\]

Answer 10

\[\large \dfrac{ \left(2\sqrt{5}\right)7 }{ \left(2\sqrt{5}\right)13 }\]

\[\large \dfrac{7}{13}\]

Answer 11

@ganeshie8 you are mistaken

Answer 12

Its okay :)

The answer would still remain the same :)

Answer 13

yes :)

Answer 14

Great work both of you !!

And sorry for being late in responding. I was busy wandering in my dream land :P

Answer 15

hehe :D no matter

Answer 16

\(\Large \sqrt{4+\sqrt{15}} +\sqrt{4-\sqrt{15}}=\sqrt{\frac{8+2\sqrt{15}}{2}} +\sqrt{\frac{8-2\sqrt{15}}{2}}\\\Large =\sqrt{\frac{(\sqrt{5}+\sqrt{3})^2}{2}}+\sqrt{\frac{(\sqrt{5}-\sqrt{3})^2}{2}}=\frac{1}{\sqrt{2}}(\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3})\\\Large =\frac{2\sqrt{5}}{\sqrt{2}}\)

\(\Large \sqrt{6+\sqrt{35}}-\sqrt{6-\sqrt{35}}=\frac{2\sqrt{5}}{\sqrt{2}}\)