Question

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

Answer 1

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)`

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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)`

Answer 2

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)`