Check whether first is a factor of second .

x+1 , x3 -1

we will get ax3 + (b+a)x2 + (c+b)x + (c+d) and after this step how we are getting the value of 'a' as 1 , 'b' as -1 and 'c' as 1

Answer :

To check  " x +  1 "  is a factor of  x³  -  1  ,  We substitute x  =  -1 in x³  -  1 and check if it gives us value as zero or not .
So, At x  = -1

= ( - 1 )³ -  1

= - 1 - 1

= - 2  , So the value is not zero , we can say that  " x +  1 "  is not a factor of  x³  -  1

And if we write ax³  + ( b  + a ) x²  + ( c  +  b ) x  + ( c  + d )

So we know from relationship between zeros and coefficient

Sum of zeros  =  0  ( From equation x³  -  1  )

So  ,

Sum of zeros =  $-\frac{\left( b +a \right)}{a}$  ( From ax³  + ( b  + a ) x²  + ( c  +  b ) x  + ( c  + d )  )

So these both are going to same as second equation we get from our equation x³  -  1 ,

So,
$-\frac{\left( b +a \right)}{a}$  = 0

- b  - a  = 0

a = - b  ----------------- ( 1 )

Sum of product of two zeros  =  0    ( From equation x³  -  1  )

So  ,

Sum of product of two zeros  =  $\frac{\left( c + b \right)}{a}$  ( From ax³  + ( b  + a ) x²  + ( c  +  b ) x  + ( c  + d )  )

So, we get

$\frac{\left( c + b \right)}{a}$  =  0

c  + b  = 0

c  = - b  --------------- ( 2 )
From equation 1 and 2 , we get

a  =  c  ----------------- ( 3 )
And

Product of all three zeros  = - ( -1 ) =  1   ( From equation x³  -  1  )

So  ,

Product of all three zeros  =  $\frac{\left( c + d \right)}{a}$  ( From ax³  + ( b  + a ) x²  + ( c  +  b ) x  + ( c  + d )  )
So, we get

$\frac{\left( c + d \right)}{a}$  =  1 

c + d  =  a  , from equation 3, we get

a  +  d  = a 

d  =  0

So we get our coefficient term is  "  (  c + d =  c + 0  =  c )

And our coefficient term in equation x³  -  1  is  -  1  ,

So,

c =  -1  ,
from equation 2

b  =  1
and
from equation 3 , we get

a  = -  1