How do you factor #64x^3-1#?

1 Answer
Jan 16, 2016

Answer:

#64x^3 - 1 = (4x-1)(16x^2+4x+1)#

#= (4x-1)(4x + 1/2 - sqrt(3)/2i)(4x + 1/2 + sqrt(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 side to verify this)

Applying this, we have:

#64x^3 - 1 = (4x)^3 - 1^3#
# = (4x-1)((4x)^2+4x*1+1^2)#
# = (4x-1)(16x^2+4x+1)#

If we wished to factor further, we could apply the quadratic formula to find the roots of #16x^2+4x+1#.

#16x^2 + 4x + 1 = 0 => x = (-4+-sqrt(4^2-4(16)(1)))/(2*16)#

#=> x = (-4+-sqrt(-48))/32#

#=> x = (-1+-sqrt(-3))/8#

As we have the square root of a negative number, we have no real solutions, and so are done if we are using only real numbers. If we allow for complex numbers, then using #i = sqrt(-1)#, we have:

#x = -1/8+-sqrt(3)/8i#

and thus, remembering to multiply by #16# to account for the coefficient of #x^2#, we can completely factor the original expression as

#64x^3 -1 = (4x-1)(4x + 1/2 - sqrt(3)/2i)(4x + 1/2 + sqrt(3)/2i)#