How do you factor $64x^3 + 343$ ?

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Answer 1 · 2017-10-31T21:30:10

$64x^3+343 = (4x+7)(16x^2-28x+49)$

Note that $64x^3 = (4x)^3$ and $343 = 7^3$ are both perfect cubes and:

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

So putting $a=4x$ and $b=7$ we find:

$64x^3+343 = (4x)^3+7^3$

$color(white)(64x^3+343) = (4x+7)((4x)^2-(4x)(7)+7^2)$

$color(white)(64x^3+343) = (4x+7)(16x^2-28x+49)$

The remaining quadratic factor has no linear factors with real coefficients. It can be factored using the primitive complex cube root of $1$ :

$omega = -1/2+sqrt(3)/2i$

Then:

$16x^2-28x+49 = (4x+7omega)(4x+7omega^2)$

How do you factor 64x^3 + 34364x^3 + 343?