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- Homework Statement
- Find the value of ##a## and ##b## in terms of ##n## given;

##(x-n+1)^3-(x-n)^3=3x^2+ax+b##

- Relevant Equations
- Binomial theorem

My approach;

Let ##(n+1)=k##

##(x-n+1)^3-(x-n)^3=3x^2+ax+b##

##(x-k)^3-(x-n)^3=3x^2+ax+b##

##(x^3-3x^2k+3xk^2-k^3)-(x^3-3x^2n+3xn^2-n^3)##

##=-3x^2k+3x^2n+3xk^2-3xn^2-k^3+n^3≡3x^2+ax+b##

##⇒1=n-k##

##⇒3(k^2-n^2)=a##

##⇒n^3-k^3=b##

##3(k+n)(k-n)=a##

##-3(k+n)=a##

##a=-3(n-1+n)##

##a=-3(2n-1)##

##a=3-6n##

##b=n^3-k^3=n^3-(n-1)^3=n^3-n^3+3n^2-3n+1=3n^2-3n+1##

##b=3n^2-3n+1##

which agrees with textbook solution...i would appreciate alternative approach.

Let ##(n+1)=k##

##(x-n+1)^3-(x-n)^3=3x^2+ax+b##

##(x-k)^3-(x-n)^3=3x^2+ax+b##

##(x^3-3x^2k+3xk^2-k^3)-(x^3-3x^2n+3xn^2-n^3)##

##=-3x^2k+3x^2n+3xk^2-3xn^2-k^3+n^3≡3x^2+ax+b##

##⇒1=n-k##

##⇒3(k^2-n^2)=a##

##⇒n^3-k^3=b##

##3(k+n)(k-n)=a##

##-3(k+n)=a##

##a=-3(n-1+n)##

##a=-3(2n-1)##

##a=3-6n##

##b=n^3-k^3=n^3-(n-1)^3=n^3-n^3+3n^2-3n+1=3n^2-3n+1##

##b=3n^2-3n+1##

which agrees with textbook solution...i would appreciate alternative approach.