Question

Simplify :
(6√2/√3+√2) - (4√3/√6+√2) + (2√6/ √2+√3)​

Answer 1

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$\huge\sf\underline\red{answered\:by\:KIMU}$

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{6 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } + \dfrac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

Consider,

$\red{\rm :\longmapsto\:\dfrac{6 \sqrt{2} }{ \sqrt{3 } + \sqrt{6} } }$

can be rewritten as

$\rm \: = \: \dfrac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } \times \dfrac{ \sqrt{6} - \sqrt{3} }{ \sqrt{6} - \sqrt{3} }$

We know that,

$\red{\boxed{ \rm{(x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

Using this identity, we get

$\rm \: = \: \dfrac{6( \sqrt{12} - \sqrt{6} )}{6 - 3}$

$\rm \: = \: \dfrac{6( \sqrt{2 \times 2 \times 3} - \sqrt{6} )}{3}$

$\rm \: = \: 2(2 \sqrt{3} - \sqrt{6)}$

$\rm \: = \: 4 \sqrt{3} - 2 \sqrt{6}$

Hence,

$\red{\rm :\longmapsto\:\dfrac{6 \sqrt{2} }{ \sqrt{3 } + \sqrt{6} } = 4 \sqrt{3} - 2 \sqrt{6} }$

Consider,

$\red{\rm :\longmapsto\:\dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \dfrac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} }$

$\rm \: = \: \dfrac{4( \sqrt{18} - \sqrt{6)} }{6 - 2}$

$\rm \: = \: \dfrac{4( \sqrt{3 \times 3 \times 2} - \sqrt{6)} }{4}$

$\rm \: = \: 3 \sqrt{2} - \sqrt{6}$

Hence,

$\red{\rm :\longmapsto\:\dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } = 3 \sqrt{2} - \sqrt{6} }$

Consider,

$\red{\rm :\longmapsto\:\dfrac{2 \sqrt{6} }{ \sqrt{2 } + \sqrt{3} } }$

can be rewritten as

$\rm \: = \: \dfrac{2 \sqrt{6} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{2 \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\rm \: = \: \dfrac{2( \sqrt{18} - \sqrt{12})}{3 - 2}$

$\rm \: = \: \dfrac{2( \sqrt{3 \times 3 \times 2} - \sqrt{2 \times 2 \times 3})}{1}$

$\rm \: = \: 2(3 \sqrt{2} - 2 \sqrt{3})$

$\rm \: = \: 6 \sqrt{2} - 4 \sqrt{3}$

Hence,

$\red{\rm :\longmapsto\:\dfrac{2 \sqrt{6} }{ \sqrt{2 } + \sqrt{3} } = 6 \sqrt{2} - 4 \sqrt{3} }$

Now, Consider,

$\purple{\bf :\longmapsto\:\dfrac{6 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } + \dfrac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } }$

$\rm \: = \: (4 \sqrt{3} - 2 \sqrt{6}) - (3 \sqrt{2} - \sqrt{6}) + (6 \sqrt{2} - 4 \sqrt{3})$

$\rm \: = \: 4 \sqrt{3} - 2 \sqrt{6} - 3 \sqrt{2} + \sqrt{6}+ 6 \sqrt{2} - 4 \sqrt{3}$

$\rm \: = \: 3 \sqrt{2} - \sqrt{6}$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)