Question

show that for every odd natural number n,

Answer 1

\[48|n^3+3n^2-n-3\]

Answer 2

n-1

Answer 3

Express the odd number n as (2N+1) for some natural number N.
Then:
\[n^3+3n^2-n-3=(2N+1)^3+3(2N+1)^2-(2N+1)-3\]
Can you continue from here or would you like me to carry on?

Answer 4

I'm not quite sure what you've done there.

Answer 5

Try expanding the right hand side and grouping terms together.

Answer 6

i substituted in the factorised form

Answer 7

\[(n-1)(n^2+3)\]

Answer 8

\[2N((2N+1)^2+3)\]\[2N(4N^2+4N+4)\]

Answer 9

\[8N(N^2+N+1)\]

Answer 10

\[48=8\times6\]

Answer 11

stuck

Answer 12

Give me a minute :)

Answer 13

okay

Answer 14

u made a little mistake somewhere it should be\[8N(N+1)(N+2)\]

Answer 15

oh i see it

Answer 16

I'm not sure, I need to take a break for a bit sorry my mind is frazzled. Maybe @Algebraic! can give you a hand?

Answer 17

thanks

Answer 18

\[8N(N+1)(N+2)\] I GET FROM THERE