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How can I derive the right side from the left side?

$n^2(n+1)^2+(n+1)^2 + n^2$ $= (n^2+n+1)^2$

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    By expanding both sides? Also, one could start with factoring out $(n+1)^2$ on the left hand side. – 2017-02-13

5 Answers 5

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$$ \begin{align} n^2(n+1)^2+(n+1)^2 + n^2 \\ & = (n(n+1))^2 +n(n+1) +(n+1) +n^2 \\ & = (n(n+1))^2 +n(n+1) +(n^2 + n) +1 \\ & = (n(n+1))^2 + 2n(n+1) +1 \\ & = (n(n+1)+1)^2 \\ & = (n^2+n+1)^2 \end{align} $$

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This may be a little late, but this is how I would start. Since we have$$(n+1)^2-n^2=2n+1\tag1$$ We now have$$n^2+(n+1)^2=2n^2+2n+1=(n^2+n+1)^2-(n^2+n)^2\tag{2}$$ Such identities arise from observations that you have occasionally. There's a cubic analogous form which I will add once I find it.

Have a nice day!

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If you compute the left side you get:

$$n^2(n^2+2n+1)+n^2+2n+1+n^2$$

so

$$n^4+2n^3+n^2+2n^2+2n+1$$

so

$$n^4+2n^3+3n^2+2n+1.$$

If you compute the right side you get:

$$(n^2+n+1)(n^2+n+1)$$

so

$$n^4+n^3+n^2+n^3+n^2+n+n^2+n+1$$

so

$$n^4+2n^3+3n^2+2n+1.$$

Finally, you can conclude that both sides are equals.

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    I was given only the left side as a small part of a larger proof where $(n^2+n+1)^2$ was the form they transformed the left side into. I was looking for a way to transform the left side into the right side, not prove that they're equal. Perhaps some factoring trick that I missed or some trinomial expansion pattern that I'm not aware of? – 2017-02-13
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set $n^2=a$ and $(n+1)=b$. then you have $(a+b)^2=a^2 +2ab + b^2$. we leave the $b^2=(n+1)^2$ alone. Then we transform $a^2 +2ab$ into $a(a+2b)=n^2(n^2+2(n+1))=n^2(n^2 +2n +1+1)=n^2[(n+1)^2 +1]$ and we end up with the lhs. I think I did the opposite of what you asked but you can see that you can invert it to get the rhs from the lhs.

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For the LHS we have: $$ n^2(n+1)^2+(n+1)^2 + n^2= $$ $$ =n^2(n^2+1+2n)+(n+1)^2+n^2=n^4+n^2+2n^3+(n+1)^2+n^2= $$ $$ =n^4+(n+1)^2+2n^2(n+1)=[n^2+(n+1)]^2 $$