Question

Rationalise the denominator 1/(√6-√3)​

Answer 1

Question :-

Rationalize the denominator :

$\longrightarrow \sf \dfrac{1}{ \sqrt{6} - \sqrt{3}}$

Answer :-

In order to rationalize the denominator, we have to multiply the numerator and denominator with the denominator's inverse such that, in the denominator :

(a-b)(a+b) = a² - b² product is formed.

$\implies \sf Denominator = \sqrt{6} - \sqrt{3}$

• a = √6
• b = √3

The inverse is √6 + √3,

multiplying the numerator and the denominator with √6 + √3,

$\implies \sf \dfrac{1}{ \sqrt{6} - \sqrt{3} } \times \dfrac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} }$

$\implies \sf \dfrac{ 1(\sqrt{6} + \sqrt{3}) }{( \sqrt{6} - \sqrt{3} )( \sqrt{6 }+ \sqrt{3} ) }$

Applying (a-b)(a+b) = a² - b² identity in the denominator,

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{ {( \sqrt{6}) }^{2} - {( \sqrt{3} )}^{2} }$

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{(6) - (3)}$

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{3}$

$\sf Therefore,\:the\:rationalized\:form \: of\:\dfrac{1}{\sqrt{6} - \sqrt{3}}\: is \: \dfrac{\sqrt{6} - \sqrt{3}}{3}$

Identities :-

• (a + b)² = a² + 2ab + b²
• (a -b)² = a² - 2ab + b²
• a² - b² = (a-b)(a+b)
• (x+a)(x+b) = x² + x(a+b) + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Conditional identity :

• If : a + b + c = 0, then a³ + b³ + c³ = 3abc

Answer 2

Step-by-step explanation:

Question :-

Rationalize the denominator :

$\longrightarrow \sf \dfrac{1}{ \sqrt{6} - \sqrt{3}}$

Answer :-

In order to rationalize the denominator, we have to multiply the numerator and denominator with the denominator's inverse such that, in the denominator :

(a-b)(a+b) = a² - b² product is formed.

$\implies \sf Denominator = \sqrt{6} - \sqrt{3}$

a = √6

b = √3

The inverse is √6 + √3,

multiplying the numerator and the denominator with √6 + √3,

$\implies \sf \dfrac{1}{ \sqrt{6} - \sqrt{3} } \times \dfrac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} }$

$\implies \sf \dfrac{ 1(\sqrt{6} + \sqrt{3}) }{( \sqrt{6} - \sqrt{3} )( \sqrt{6 }+ \sqrt{3} ) }$

Applying (a-b)(a+b) = a² - b² identity in the denominator,

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{ {( \sqrt{6}) }^{2} - {( \sqrt{3} )}^{2} }$

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{(6) - (3)}$

$\implies \sf \dfrac{ \sqrt{6} + \sqrt{3} }{3}$

$\sf Therefore,\:the\:rationalized\:form \: of\:\dfrac{1}{\sqrt{6} - \sqrt{3}}\: is \: \dfrac{\sqrt{6} - \sqrt{3}}{3}$

Identities :-

(a + b)² = a² + 2ab + b²

(a -b)² = a² - 2ab + b²

a² - b² = (a-b)(a+b)

(x+a)(x+b) = x² + x(a+b) + ab

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Conditional identity :

If : a + b + c = 0, then a³ + b³ + c³ = 3abc