Question

6+√3/6-√3=a+b√3 answer pls​

Answer 1

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Answer 2

Answer:

Required value of a is $\frac{13}{11}$ and required value of b is $\frac{4}{11}$

Step-by-step explanation:

Given,

$\frac{6 + \sqrt{3} }{6 - \sqrt{3} } = a + b \sqrt{3} \\ a + b \sqrt{3} = \frac{(6 + \sqrt{3})(6 + \sqrt{3})}{(6 - \sqrt{3})(6 + \sqrt{3})} \\ a + b \sqrt{3} = \frac{ {(6 + \sqrt{3})}^{2} }{ {6}^{2} - { \sqrt{3} }^{2} } \\ a + b \sqrt{3} = \frac{36 + 12 \sqrt{3} + 3 }{36 - 3} \\ a + \sqrt{3} b = \frac{39 + 12 \sqrt{3} }{33} \\ a + \sqrt{3} b = \frac{13 + 4 \sqrt{3} }{11} \\ a + \sqrt{3} b = \frac{13}{11} + \frac{4}{11} \sqrt{3}$

We are comparing both side,

$a = \frac{13}{11}$ and $b = \frac{4}{11}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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