How do you factor 64r^3 - 27?

Feb 27, 2016

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

$64 {r}^{3} - 27 = \left(4 r - 3\right) \left(16 {r}^{2} + 12 r + 9\right)$

$= \left(4 r - 3\right) \left(4 r + \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(4 r + \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

Explanation:

The difference of cubes formula states that
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
(Try multiplying the right hand side to verify this).

In this case, that gives us

$64 {r}^{3} - 27 = {\left(4 r\right)}^{3} - {3}^{3}$

$= \left(4 r - 3\right) \left(16 {r}^{2} + 12 r + 9\right)$

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.

$16 {r}^{2} + 12 r + 9 = 0$

$\iff r = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \left(16\right) \left(9\right)}}{2 \left(16\right)}$

$= \frac{- 12 \pm \sqrt{- 432}}{32}$

$= \frac{- 12 \pm 12 \sqrt{3} i}{32}$

$= - \frac{3}{8} \pm \frac{3 \sqrt{3}}{8} i$

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

$64 {r}^{3} - 27 =$
$= \left(4 r - 3\right) \left(4 r + \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(4 r + \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

Feb 27, 2016

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

$\left({x}^{3} - {y}^{3}\right) = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$.

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

= $\left(4 r - 3\right) \left(16 {r}^{2} + 12 r + 9\right)$

It is obvious that this cannot be factorized further, as the determinant ${12}^{2} - 4 \times 16 \times 9 = - 432$, is negative.

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