If#a+b+c=0#,show that,#(a+bomega+comega^2)^3+(a+bomega^2+comega)^3=27abc#?

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Answer 1 · 2018-05-29T10:03:07

# a + b + c = 0#

#LHS = (a + bω + cω²)³ + (a + bω² + cω)³ #

If we consider #A = (a + bω + cω²)# and #B = (a + bω² + cω)#

Adding the two.

#color(magenta)(A + B) = 2a + b(omega+omega^2) + c(omega^2 +omega)#
#=> 2a-b-c = 3a-a-b-c = 3a-0 = color(magenta)(3a)# #"as " [ω² + ω + 1 = 0]#

Multiplying the two

#color(magenta)(AB) = (a + b + c)² - 3(ab + bc + ca) = color(magenta)(- 3(ab + bc + ca) #

As we know, #(A+B)^3 = A^3+B^3 +3AB(A+B)#

#P => A^3+B^3 = (A + B)³ - 3AB(A + B)#
# = 27a³ + 27a(ab + bc + ca) #
#= 27a(a² + ab + ac + bc) = 27a(a + b)(a + c) = 27a(-c)(-b) = 27abc#