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)#