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!
The left-hand side Can be written as $$(m+1)\left(\frac{1}{4}m(m^2+m+2)+(m+1)^2+1\right)$$ and this is equal to $$(m+1)\left(\frac{m(m^2+m+2)+4(m+1)^2+4}{4}\right)$$ and this is equal to $$\frac{(m+1)(m+2)(m^2+3m+4)}{4}$$