How do you factor #64r^3 - 27#?

2 Answers
Feb 27, 2016

Answer:

Use the difference of cubes formula (and the quadratic formula if you allow for complex numbers) to find that

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

#=(4r-3)(4r+3/2-(3sqrt(3))/2i)(4r+3/2+(3sqrt(3))/2i)#

Explanation:

The difference of cubes formula states that
#a^3-b^3 = (a-b)(a^2+ab+b^2)#
(Try multiplying the right hand side to verify this).

In this case, that gives us

#64r^3-27 = (4r)^3-3^3#

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


If we are only consider real numbers, then we are done. If we are allowing for complex numbers, then we can continue factoring using the quadratic formula on the remaining quadratic expression.

#16r^2+12r+9 = 0#

#<=>r = (-12+-sqrt(12^2-4(16)(9)))/(2(16))#

#=(-12+-sqrt(-432))/32#

#=(-12+-12sqrt(3)i)/32#

#=-3/8+-(3sqrt(3))/8i#

Thus, making sure to multiply by #16# to obtain the correct coefficients, the complete factorization would be

#64r^3-27 =#
#=(4r-3)(4r+3/2-(3sqrt(3))/2i)(4r+3/2+(3sqrt(3))/2i)#

Feb 27, 2016

Answer:

Hence factors are given by #(64r^3−27)=(4r-3)(16r^2+12r+9)#

Explanation:

As the function #64r^3−27=(4r)^3-3^3#, one could use the identity

#(x^3-y^3)=(x-y)(x^2+xy+y^2)#.

Using this, #64r^3−27=(4r-3)((4r)^2+4rxx3+3^2)#

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

It is obvious that this cannot be factorized further, as the determinant #12^2-4xx16xx9=-432#, is negative.

Hence factors are given by #(64r^3−27)=(4r-3)(16r^2+12r+9)#