Math, asked by elsa8383, 1 year ago

At which value in the domain does mc027-1.jpg? mc027-2.jpg mc027-3.jpg mc027-4.jpg mc027-5.jpg mc027-6.jpg

Answers

Answered by shoaibahmad131
0

Answer:

1,2,3,4,5,6 is in domain. mc027 is fix.

Step-by-step explanation:

solve equation

mc27-1jpg  

Final result :

 mc27 - jpg

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "c0"   was replaced by   "c^0".  

(2): Dot was discarded near "1.j".

Step by step solution :

Step  1  :

Trying to factor as a Difference of Cubes:

1.1      Factoring:  mc27-jpg  

Theory : A difference of two perfect cubes,  a3 - b3 can be factored into

             (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =

           a3+a2b+ab2-ba2-b2a-b3 =

           a3+(a2b-ba2)+(ab2-b2a)-b3 =

           a3+0+0+b3 =

           a3+b3

Check :  m 1 is not a cube !!  

Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

 mc27 - jpg

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