Question

$\frac{ {x}^{2} - 9 }{x + 2} + \frac{ {x}^{3} - 27 }{x + 2}$


को परिमेय व्यंजक के रूप में व्यक्त कीजिए

Answer 1

$\bold{\huge{Hay!!}}$


$\bold{Dear\:user!!}$



$\bold{\underline{Question-}}$



$\frac{ {x}^{2} - 9 }{x + 2} + \frac{ {x}^{3} - 27 }{x + 2}$

को परिमेय व्यंजक के रूप में व्यक्त कीजिए





$\bold{\underline{Answer-}}$

$\frac{ {x}^{2} - 9 }{x + 2} + \frac{ {x}^{3} - 27 }{x + 2}$


$= > \frac{ {x}^{2} - 9 + {x}^{3} - 27 }{(x + 2)} \\ \\ \\ \\ = > \frac{ {x}^{3} + {x}^{2} - 36 }{x + 2}$

Answer 2

HEY Buddy.....!! here is ur answer

Given that :

$\frac{ {x}^{2} - 9 }{(x + 2)} + \frac{ {x}^{3} - 27}{(x + 2)} \\ \\ = > \frac{ {(x)}^{2} - {(3)}^{2} }{(x + 2)} + \frac{ {(x)}^{3} - {(3)}^{3} }{(x + 2)} \\ \\ As \: we \: know \: that: \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \: and \: {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ = > \frac{(x + 3)(x - 3)}{(x + 2)} + \frac{(x - 3)( {x}^{2} + 3x + {9}) }{(x + 2)} \\ \\ = > \frac{(x + 3)(x - 3) + (x - 3)( {x}^{2} + 3x + 9) }{(x + 2)} \\ \\ = > \frac{(x - 3)(x + 3 + {x}^{2} + 3x + 9) }{(x + 2)} \\ \\ = > \frac{(x - 3)( {x}^{2} + 4x + 12) }{(x + 2)}$

I hope it will be helpful for you...!!

THANK YOU ✌️✌️