Question

if x= √7+2√12 and y=√7-2√12 find x^3+y^3​

Answer 1

Answer:

$\large\bold\red{ {x}^{3} + {y}^{3} = 302 \sqrt{7} }$

Step-by-step explanation:

Given,

$x = \sqrt{7} + 2 \sqrt{12}$

And,

$y = \sqrt{7} - 2 \sqrt{12}$

Now,

We know that,

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

Therefore,

We get,

$= > x + y = \sqrt{7} + 2 \sqrt{12} + \sqrt{7} - 2 \sqrt{12} \\ \\ = > x + y = 2 \sqrt{7}$

And,

$= > {x}^{2} = {( \sqrt{7} + 2 \sqrt{12} )}^{2} \\ \\ = > {x}^{2} = {( \sqrt{7}) }^{2} + {(2 \sqrt{12}) }^{2} + 2 \times \sqrt{7} \times 2 \sqrt{12} \\ \\ = > {x}^{2} = 7 + 48 + 8 \sqrt{21} \\ \\ = > {x}^{2} = 55 + 8 \sqrt{21}$

Similarly,

We have,

$= > {y}^{2} = {( \sqrt{7} - 2 \sqrt{12}) }^{2} \\ \\ = > {y}^{2} = 55 - 8 \sqrt{21}$

Also,

$= > xy = ( \sqrt{7} + 2 \sqrt{12} )( \sqrt{7} - 2 \sqrt{12} ) \\ \\ = > xy = {( \sqrt{7}) }^{2} - {(2 \sqrt{12} )}^{2} \\ \\ = > xy = 7 - 48 \\ \\ = > xy = - 41$

Now,

Substituting the respective values,

We get,

$= > {x}^{3} + {y}^{3} = 2 \sqrt{7}(55 + 8 \sqrt{21} + 55 - 8 \sqrt{21} - ( - 41)) \\ \\ = > {x}^{3} + {y}^{3} = 2 \sqrt{7} (110 + 41) \\ \\ = > {x}^{3} + {y}^{3} =151 \times 2 \sqrt{7} \\ \\ = > \bold{ {x}^{3} + {y}^{3} = 302 \sqrt{7} }$