If 1,w,w^2 denotes the cube root of unity ,find the roots of (x+5)^3+27=0?

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Answer 1 · 2017-12-29T11:14:26

The given equation

# (x+5)^3+27=0#

# =>(x+5)^3=-27#

# =>(x+5)^3=-3^3#

# =>((x+5)/-3)^3=1#

Let #((x+5)/-3)=y#

So we have

#y^3=1#
As #1.omega,omega^2# are the cube roots of unity then we have

#y =1 ,y=omega and y=omega^2#

Hence we get
when #y =1 #
#((x+5)/-3)=1#

#=>x=-3-5=-8#

when #y =omega #

#((x+5)/-3)=omega#

#=>x=-3omega-5#

when #y =omega^2 #

#((x+5)/-3)=omega^2#

#=>x=-3omega^2-5#

So three roots of the given equations are

#-8,(-3omega-5),(-3omega^2-5)#