An algebra problem by Aaryan Maheshwari

Relevant wiki: Algebraic Identities

$\text{We have been given:}\space (a^2-bc)^2-(b^2-ca)(c^2-ab).$

$\text{Simplifying the above expression, we get:}$

$a^4+ab^3+ac^3-3a^2bc\space =\space a(a^3+b^3+c^3-3abc)$

$\text{Now, we will prove that if}\space a+b+c=0, \text{we have}\space a^3+b^3+c^3=3abc.$

$\text{Since we know that}\space a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca), \text{we have:}$

$a^3+b^3+c^3-3abc= 0\times(a^2+b^2+c^2-ab-bc-ca)=0. \text{Thus,}$

$\therefore\space a(a^3+b^3+c^3-3abc)=a\times0=\boxed{0}.$