Question

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

Answer 1

Step-by-step explanation:

LHS=acu + bcu + ccu -3abc

RHS = acu +bcu+ccu-3abc

Answer 2

卂几丂山乇尺:

first of all you should be know that

first of all you should be know that formula

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

LHS = a³ + b³ + c³ - 3abc

= ( a³ + b³ ) + c³ - 3abc

= ( a + b)³ - 3ab( a + b) +c³ - 3abc

= ( a + b)³ - 3a²b - 3ab² + c³ - 3abc

= {(a + b)³ + c³ } -3a²b - 3ab² - 3abc

= ( a + b + c)³ - 3c(a + b) ( a + b + c ) -3ab(a + b+ c)

= ( a + b + c){ ( a + b + c)² -3c(a +b) - 3ab }

= ( a + b + c){ a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab }

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

LHS = RHS

::::::::::::::Hence, Proved:::::::::::