Question

Find the factors of ${(a+b+c)}^5-{(b+c-a)}^5-{(c+a-b)}^5-{(a+b-c)}^5$.

Answer 1

In order to find the factors of $(a+b+c{)}^5-(b+c-a{)}^5-(c+a-b{)}^5-(a+b-c{)}^5$ we have to expand each term by binomial theorem

Let us consider

$\left[\right(a+b+c{)}^5-(b+c-a{)}^5]-\left[\right(c+a-b{)}^5+(a+b-c{)}^5]$

Expansion of each term separately,

$(a+b+c{)}^5=$ $a^5+5a^{4}b+10a^3{b}^2+10a^2{b}^3+5a{b}^4+b^5+5a^{4}c+20a^{3}bc+30a^2{b}^{2}c+20a{b}^{3}c+5b^{4}c+10a^3{c}^2+30a^{2}b{c}^2$

$+30a{b}^2{c}^2+10b^3{c}^2+10a^2{c}^3+20ab{c}^3+10b^2{c}^3+5a{c}^4+5b{c}^4+c^5$

$(b+c-a{)}^5=$$-a^5+5a^{4}b+5a^{4}c-10a^3{b}^2-20a^{3}bc-10a^3{c}^2+10a^2{b}^3+30a^2{b}^{2}c+30a^{2}b{c}^2+10a^2{c}^3-5a{b}^4-20a{b}^{3}c-30a{b}^2{c}^2$

$-20ab{c}^3-5a{c}^4+b^5+5b^{4}c+10b^3{c}^2+10b^2{c}^3+5b{c}^4+c^5$

$(c+a-b{)}^5=$ $a^5-5a^{4}b+5a^{4}c+10a^3{b}^2-20a^{3}bc+10a^3{c}^2-10a^2{b}^3+30a^2{b}^{2}c-30a^{2}b{c}^2+10a^2{c}^3+5a{b}^4-20a{b}^{3}c$

$+30a{b}^2{c}^2-20ab{c}^3+5a{c}^4-b^5+5b^{4}c-10b^3{c}^2+10b^2{c}^3-5b{c}^4+c^5$

$(a+b-c{)}^5=$ $a^5+5a^{4}b-5a^{4}c+10a^3{b}^2-20a^{3}bc+10a^3{c}^2+10a^2{b}^3-30a^2{b}^{2}c+30a^{2}b{c}^2-10a^2{c}^3+5a{b}^4-20a{b}^{3}c$

$+30a{b}^2{c}^2-20ab{c}^3+5a{c}^4+b^5-5b^{4}c+10b^3{c}^2-10b^2{c}^3+5b{c}^4-c^5$

Now,

$(a+b+c{)}^5-(b+c-a{)}^5=$ $2a^5+20a^3{b}^2+40a^{3}bc+20a^3{c}^2+10a{b}^4+40a{b}^{3}c+60a{b}^2{c}^2+40ab{c}^3+10a{c}^4$

$(c+a-b{)}^5+(a+b-c{)}^5=$ $2a^5+20a^3{b}^2-40a^{3}bc+20a^3{c}^2+10a{b}^4-40a{b}^{3}c+60a{b}^2{c}^2-40ab{c}^3+10a{c}^4$

$\therefore$ $\left[\right(a+b+c{)}^5-(b+c-a{)}^5]-\left[\right(c+a-b{)}^5+(a+b-c{)}^5]$

$=$ $[2a^5+20a^3{b}^2+40a^{3}bc+20a^3{c}^2+10a{b}^4+40a{b}^{3}c+60a{b}^2{c}^2+40ab{c}^3+10a{c}^4]$ $-$

$[2a^5+20a^3{b}^2-40a^{3}bc+20a^3{c}^2+10a{b}^4-40a{b}^{3}c+60a{b}^2{c}^2-40ab{c}^3+10a{c}^4]$

$=$ $80a{b}^{3}c+80ab{c}^3+80a^{3}bc$

$=80abc(a^2+b^2+c^2)$

$\therefore$ Factors of $(a+b+c{)}^5-(b+c-a{)}^5-(c+a-b{)}^5-(a+b-c{)}^5$ is $80abc(a^2+b^2+c^2)$.