How do you factor $27 k^{3} + 64 d^{3}$ $27k^3+64d^3$ ?

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How do you factor $27 k^{3} + 64 d^{3}$ $27k^3+64d^3$ ?

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Answer 1 · 2016-08-29T08:50:56

$27 k^{3} + 64 d^{3} = (3 k + 4 d) (9 k^{2} - 12 k d + 16 d^{2})$ $27k^3+64d^3=(3k+4d)(9k^2-12kd+16d^2)$

Explanation:

The sum of cubes identity can be written:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$

We can use this with $a = 3 k$ $a=3k$ and $b = 4 d$ $b=4d$ to find:

$27 k^{3} + 64 d^{3}$ $27k^3+64d^3$

$= {(3 k)}^{3} + {(4 d)}^{3}$ $=(3k)^3+(4d)^3$

$= (3 k + 4 d) ({(3 k)}^{2} - (3 k) (4 d) + {(4 d)}^{2})$ $=(3k+4d)((3k)^2-(3k)(4d)+(4d)^2)$

$= (3 k + 4 d) (9 k^{2} - 12 k d + 16 d^{2})$ $=(3k+4d)(9k^2-12kd+16d^2)$

That is as far as we can go with Real coefficients. If we allow Complex coefficients then this can be factored further as:

$= (3 k + 4 d) (3 k + 4 ω d) (3 k + 4 ω^{2} d)$ $=(3k+4d)(3k+4omegad)(3k+4omega^2d)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$

Answer 2 · 2016-08-29T09:07:02

= $(3 k + 4 d) (9 k^{2} - 12 d k + 16 d^{2})$ $(3k +4d)(9k^2 -12dk +16d^2)$

Explanation:

This difficulty is recognising that the values are all cubes.

This expression is the sum of cubes.

$x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ $x^3 + y^3 = (x+y)(x^2 -xy +y^2)$

$27 k^{3} + 64 d^{3} = (3 k + 4 d) (9 k^{2} - 12 d k + 16 d^{2})$ $27k^3 +64d^3 = (3k +4d)(9k^2 -12dk +16d^2)$

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How do you factor $27 k^{3} + 64 d^{3}$ $27k^3+64d^3$ ?

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How do you factor $27 k^{3} + 64 d^{3}$ $27k^3+64d^3$ ?