I have been trying to proof the following for more than a day, but to no avail. I have tried several ways, partial/full expansion/factorisation etc but I just cannot prove it. I would love indirect nudges and hints, rather than direct answers since I wish to work it out myself, as I'm sure many do prefer of the questioner.
I am supposed to prove, starting from LHS that
$\frac{1}{4}(m)(m+1)(m^2+m+2)+(m+1)[(m+1)^2+1]=\frac{1}{4}(m+1)(m+2)(m^2+3m+4)$
I zoomed in on the the fact that $\frac{1}{4}$ and $(m+1)$ can be factorised i.e.
$LHS = \frac{1}{4}(m)(m+1)(m^2+m+2)+(m+1)[(m+1)^2+1]$
= $\frac{1}{4}(m+1)[m(m^2+m+2)+4(m^2+2m+2)]$
Seeing that $(m+2)$ was also factorised on the RHS (what I'm working towards), I sought to also factorise $(m+2)$ out. I mistakenly thought I saw an opportunity to do this:
= $\frac{1}{4}(m+1)[m(m^2+m+2)+2(2)(m^2+2m+2)]$
= $\frac{1}{4}(m+1)[m(m^2+m+2)+2(2m^2+4m+4)]$
And now, unfortunately, I am unable to factorise because $(m^2+m+2)\not= (2m^2+4m+4)$. In any case, none of them are equal to the expression $(m^2+3m+4)$ that I'm trying to obtain (see RHS).
I feel like I'm missing something really obvious yet crucial here, some technique of manipulation or something. Would really appreciate any hints or nudges in the right direction. Thank you!
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