Question

Prove the following identities:
$abc(\sum a{)}^3-(\sum bc{)}^3=abc\sum a^3-\sum b^3{c}^2=(a^2-bc)(b^2-ca)(c^2-ab)$.

Answer 1

Let $I_1=abc{(\sum a)}^3-{(\sum bc)}^3$

$I_2=abc\sum a^3-\sum b^3{c}^3$

$I_3=(a^2-bc)(b^2-ca)(c^2-ab)$

Therefore,

$I_1=abc{(\sum a)}^3-{(\sum bc)}^3$

$=abc{(a+b+c)}^3-{(ab+bc+ca)}^3$

$=abc(a^3+b^3+c^3+3a^{2}b+3a{b}^2+3b^{2}c+3b{c}^2+3c^{2}a+3c{a}^2+6abc)-$

$({\left(ab\right)}^3+{\left(bc\right)}^3+{\left(ca\right)}^3+3a^3{b}^{2}c+3a^{3}b{c}^2+3a{b}^3{c}^2+3a{b}^2{c}^3+3a^2{b}^{3}c+3a^{2}b{c}^3+3a^2{b}^2{c}^2)$

$=abc(a^3+b^3+c^3)-({\left(ab\right)}^3+{\left(bc\right)}^3+{\left(ca\right)}^3)+3abc(b^{2}c+b{c}^2+a^{2}b+a{b}^2+a{c}^2+a^{2}c+2abc)$

$-3abc(b^{2}c+b{c}^2+a^{2}b+a{b}^2+a{c}^2+a^{2}c+2abc)$

$=abc(a^3+b^3+c^3)-({\left(ab\right)}^3+{\left(bc\right)}^3+{\left(ca\right)}^3)$

$=I_2$

$=a^{4}bc+a{b}^{4}c+ab{c}^4-a^3{b}^3-b^3{c}^3-c^3{a}^3$

$=a^{4}bc-b^3{c}^3-a^3{b}^3+a{b}^{4}c+ab{c}^4-c^3$

$=bc(a^4-b^2{c}^2)-a{b}^3(a^2-bc)-a{c}^3(a^2-bc)$

$=(a^2-bc)(bc(a^2+bc)-a{b}^3-a{c}^3)$

$=(a^2-bc)(b^2-ac)(c^2-ab)$

$=I_3$