Question

a³ - b³ - c³ - 3abc =

Answer 1

Hi ,

We know that ,
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An algebraic identity ,

x³ + y³ + z³ - 3xyz

= ( x+y+z)(x²+y²+z² -xy -yz -zx )

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Here ,

x = a , y = - b , z = -c

a³ + ( -b)³ + (-c)³ -3a(-b)(-c)

= (a-b-c)[ a² + (-b)²+(-c)²-a(-b)-(-b)(-c)-
(-c)a ]

Therefore ,

a³ - b³ - c³ - 3abc

=( a- b - c)[ a²+b²+c²+ab -bc + ca ]

I hope this helps you.

:)

Answer 2

a³➕b³➕c³=3abc

↔a³➕b³➕c³➖3abc=0

↔(a➕b➕c)✖(a²➕b²➕c²➖ab➖bc➖ca)=0