Question

show that a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)

Answer 1

(a+b+c)(a2+b2+c2-ab-bc-ca) 

= a^3+ab^2+ac^2-a^2b-abc-ca^2 + ba^2+b^3+bc^2-ab^2-b^2c-bca + ca^2+cb^2+c^3-cab-bc^2-c^2a 

= (a^3+b^3+c^3) + (ab^2-ab^2) + (ac^2-c^2a) + (-a^2b+ba^2) + (-abc-bca-cab) + (-ca^2+ca^2) + (bc^2-bc^2) + (-b^2c+c^2b) 

= a^3+b^3+c^3 -3abc

Answer 2

( a + b + ) ( a² + b² + c² - ab - bc - ca )

= a ( a² + b² + c² - ab - bc - ca ) + b ( a² + b² + c² - ab - bc - ca ) + c ( a² + b² + c² - ab - bc - ca )

= a³ + ab² + ac² - a²b - abc - ca² + a²b + b³ + bc² - ab² + b²c - abc + ca² + cb² + c³ - abc - bc² - c²a

= a³ + b³ + c³ + ab² - ab² + ac² - ac² + a²b - a²b + a²c - a²c + bc² - bc² + b²c - b²c - abc - abc - abc

= a³ + b³ + c³ - 3abc

This , we have the following identity :

a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )