Question

prove that the difference of cubes of two consecutive number is one more than thrice the product of tje number algebraically.Is it true of negative integers also . Justify the answer with example .PLEASE HELP ME TO DO THIS QUESTION

Answer 1

let consecutive numbers be x and x+1
so (x+1)^3-x^3=x^3+1+3x^2+3x-x^3
=3x^2+3x+1
=3*x*(x+1) +1
=thrice the product plus 1

Answer 2

Answer:

let consecutive numbers be x and x+1

so (x+1)^3-x^3=x^3+1+3x^2+3x-x^3

=3x^2+3x+1

=3*x*(x+1) +1

=thrice the product plus 1

Step-by-step explanation: