How do you solve #64x^3+27=0#?#?

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Answer 1 · 2015-09-29T23:24:51

Factor #64x^3+27 = (4x+3)(16x^2-12x+9)#

Hence #64x^3 + 27 = 0# when #4x+3 = 0#, that is when #x = -3/4#

In general #a^3+b^3 = (a+b)(a^2-ab+b^2)#

Putting #a=4x# and #b=3# we find:

#64x^3+27 = (4x)^3+3^3 = (4x + 3)((4x)^2 - (4x)*3 + 3^2)#

#= (4x+3)(16x^2 - 12x + 9)#

The remaining quadratic factor has no linear factors with Real coefficients.

You can tell that because its discriminant is negative:

#16x^2 - 12x + 9# is of the form #ax^2+bx+c# with #a = 16#, #b = -12# and #c = 9#.

Its discriminant #Delta# is given by the formula:

#Delta = b^2 - 4ac = (-12)^2 - (4xx16xx9) = 144 - 576#

#= -432#

In fact, #16x^2 - 12x + 9 = (4x+3omega)(4x+3omega^2)#

where #omega = -1/2 + sqrt(3)/2i# is the primitive cube root of unity.

#omega# is used when finding all the roots of a general cubic equation.

Answer 2 · 2015-09-30T01:07:19

#x= -3/4#

(As an alternate method)
#64x^3+27=0#

#rArr (4x)^3 = (-3)^3#

#rArr 4x=-3#

#rArr x= -3/4#