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$[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] +6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$

After the simplification , the result will be:

$3pq \cdot (p^2q + pq^2+2p^2-2pq+2q^2)$

I have tried to simplify it by using the binomial expansion for the power terms and by multiplying , adding, subtracting. it takes a lot of time.

Is there any alternative easy way to simplify the following term?

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    $\begingroup$ @Carl Najafi Thank you very much for the additional Tag (algebra-precalculus) & also reforming the question. $\endgroup$ – time May 23 '13 at 6:09
  • $\begingroup$ WA does it $\endgroup$ – lhf Jul 1 at 0:56
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Well... This bizzare thing $$[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] +6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$$

Can be split like : $T_1-T_2-3T_2+3T_3+3T_3-3T_4$ , Where $T_i$ denotes the $i^{\text{th}}$ term.

$[p (p+1) (p+2) (p+3) (p+q)^3] - [ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] -3[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] +3[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)]+ 3[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$

Now take the common terms out, $$p(p+1)(p+2)(p+q)^2[(p+3)(p+q)-p(p+q+3)] \\ 3p^2(p+1)(p+q)(p+q+3)[p(p+q+2)-(p+3)(p+q)] \\ + 3p^3(p+q+2)(p+q+3)[(p+1)(p+q)-p(p+q+1)]$$

Consider only the terms inside the square brackets for now. It's the best to just expand them as it is $$ (p+3)(p+q)-p(p+q+3) = p^2 +pq +3p+3q-p^2 -pq-3p =3q \\ p(p+q+2)-(p+3)(p+q)= p^2+pq+2p -p^2 -pq-2p-2q= -2q \\ (p+1)(p+q)-p(p+q+1)= p^2+pq+p+q-p^2-pq-p =q$$

Now the equation becomes: $$3pq(p+1)(p+2)(p+q)^2-6p^2q(p+1)(p+q)(p+q+3) +3p^3q(p+q+2)(p+q+3)$$ Again, take the common temrs out, i.e. $3pq$

$$3pq[(p+1)(p+2)(p+q)^2-2p(p+1)(p+q)(p+q+3) +p^2(p+q+2)(p+q+3)$$

Again, split the middle term and common terms out: $$3pq[(p+1)(p+q)[(p+2)(p+q)-p(p+q+3)]+p^2(p+q+3)[p(p+q+2)-(p+1)(p+q)]]$$

Again, expand the terms in inner square brackets: $$(p+2)(p+q)-p(p+q+3)=p^2+2p+2q+pq-p^2-pq-3p= 2q-p \\ p(p+q+2)-(p+1)(p+q)= p^2+pq+2p-p^2-pq-p-q=p-q $$ The equation becomes: $$3pq[(p+1)(p+q)(2q-p)+p^2(p-q)(p+q+3)]$$

I don't think it can be simplified more, so it's best to expand that... $$=3pq[2p^2q+2pq^2+2pq+2q^2-p^3-p^2q-p^2-pq +p^3+p^2q+3p^2-p^2q-pq^2-3pq]$$

Which is indeed equal to $$3pq[p^2q+pq^2-2pq+2p^2+2q^2]$$

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