- #1
chwala
Gold Member
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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.