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Toolkit 18

Symmetric cubic sum

a3+b3+c3=(a+b+c)(a2+b2+c2−ab−ac−bc)+3abca^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc

Proof

Expanding the right-hand side gives

(a+b+c)(a2+b2+c2−ab−ac−bc)+3abc=a3+ab2+ac2−a2b−a2c−abc+a2b+b3+bc2−ab2−abc−b2c+a2c+b2c+c3−abc−ac2−bc2+3abc.\begin{aligned} &(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc \\ &= a^3+ab^2+ac^2-a^2b-a^2c-abc \\ &\quad + a^2b+b^3+bc^2-ab^2-abc-b^2c \\ &\quad + a^2c+b^2c+c^3-abc-ac^2-bc^2+3abc. \end{aligned}

All mixed terms cancel, leaving

a3+b3+c3.□a^3+b^3+c^3. \quad\square