Question

plz answere this fast
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Answer 1

Answer:

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:ARADHYA}}$

SOLUTION:

GIVEN EXPRESSION:

$x^{3}+3x^{2}-x-3$

On Factorising, we have :

$x^{2}(x+3) -1(x+3)$

=$(x^{2}-1) (x+3)$

= $(x+1) (x-1) (x+3)$

$\large\red{\boxed{a^{2}-b^{2}=(a+b) (a-b) }}$

On equating with zero.

$(x+1) (x-1) (x+3) =0$

Therefore x = -1, 1,-3

So $\alpha$ = -1

$\beta$=1

$\gamma$=-3

So, $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$+$\dfrac{1}{\gamma}$

= $\dfrac{1}{-1}+ \dfrac{1}{1} + \dfrac{1}{-3}$

=$1-1-\dfrac{1}{3}$

=$\dfrac{-1}{3}$