How do you solve #(x-1)^3+8 =0 # ?

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Answer 1 · 2017-09-18T09:29:59

#x= -1 or 2+-sqrt3 i#

#(x-1)^3+8 =0 -> (x-1)^3 =-8#

Let #lambda = x-1 -> x=lambda+1#

#:. lambda^3=-8#

#lambda =root3 (-8)#

#lambda = -2 -> x=-2+1 = -1# is a root of the original equation

and #(lambda +2)# is a factor of #(lambda^3+8)#

Now consider: #(lambda^3+8)/(lambda+2) = lambda^2-2lambda+4#

Hence, #lambda^2-2lambda+4=0# will yield the remaining roots.

#lambda = (+2+-sqrt((-2)^2-4*1*4))/(2*1)#

#= (2+-sqrt(4-16))/2 = 1+-sqrt(-12)/2#

#=1+-(2sqrt(-3))/2 = 1+-sqrt3 i#

Since #x=lambda+1 -> x=2+-sqrt3 i# are the complex roots of the original equation.

#:. x= -1 or 2+-sqrt3 i# are the three roots of the cubic equation.