Math Induction: Solving (n^3+5n)=6q

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Homework Statement



n [tex]\epsilon[/tex][tex]N[/tex]=n[tex]\geq0[/tex]
6 divides (n[tex]^{3}[/tex]+5n)

Homework Equations


(n[tex]^{3}[/tex]+5n)=6q


The Attempt at a Solution


by expanding and simplifying and later on substituting 6q in
(n+1)[tex]^{}3[/tex]+5(n+1)

ive arrived at 6q+3n[tex]^{}2[/tex] +3n+6

then.. I am stuck...pls help...lot of thanks!
 
on Phys.org
i tried multiplying and dividing it by two so that the 6 would be factored out...ive got fractions..its supposed to be whole numbers
 
(n+1)^3 +5(n+1)
n^3 + 3n^2 + 3n + 1 + 5n + 5
(n^3 + 5n) + 3n^2 + 3n + 6
6q + 3n^2 + 3n + 6
This reduces the problem down to proving that 3n^2 + 3n + 6 is a multiple of 6

3(n^2 + n + 2)
Now you just have to prove that n^2 + n + 2 is a multiple of 2, (is even).
 
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