What is the Manipulation of Algebraic Expressions?

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Homework Statement



I need some help with this Algebra problem. In the following I don't know how they manipulated the RHS to get to the LHS:

http://img402.imageshack.us/img402/7176/solns.gif

Homework Equations



The Attempt at a Solution



[tex]\frac{1}{4}(k+1)^2 .k^2 + (k+1)^3[/tex]

[tex]\frac{1}{4}(k+1)^2 .k^2 + k^3 + 3k^2 +3k +1[/tex]

I don't know how to manipulate [tex]k^2 + k^3 + 3k^2 +3k +1[/tex] into [tex](k^2+4k+4)[/tex]
 
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Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.
 
another opinion of mine personally is to resolve 'complex' question initially is to model them by substitution.
Like letting another alegrabic representation (eg) Let a = K+1 and then it will look simpler.

Hope it helps.
 
symbolipoint said:
Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.

Well I'm not sure what they've done there. I mean how they simplified [tex][\frac{1}{2}k(k+1)]^2 + (k+1)^3[/tex] into [tex]\frac{1}{4} (k+1)^2 (k^2+4k+4)[/tex]

Anyway, here's the rest (it's from a proof by induction problem):

http://img200.imageshack.us/img200/6978/62815040.gif
 
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[tex][\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2[/tex]

Now just factor (k+1)2 and you are done. :approve:
 
njama said:
[tex][\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2[/tex]

Now just factor (k+1)2 and you are done. :approve:

But that doesn't work! If I factor out the (k+1)2 I will have:

[tex]\frac{1}{4}(k+1)^2 (k^2+k+1)[/tex]

Which is not the same as:

[tex]\frac{1}{4} (k+1)^2 (k^2+4k+4)[/tex]

:rolleyes:
 
roam said:
But that doesn't work! If I factor out the (k+1)2 I will have:

[tex]\frac{1}{4}(k+1)^2 (k^2+k+1)[/tex]

Which is not the same as:

[tex]\frac{1}{4} (k+1)^2 (k^2+4k+4)[/tex]

:rolleyes:

You are wrong.

Do the factorization again.

[tex]=(k+1)^2(\frac{1}{4}k^2+k+1)[/tex]

Now factor 1/4 and see what will you come up with.
 
Awww! Gee! I see what you mean now! Thanks a lot for the help.