Question 179786
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This is the sum of two cubes.  64 is 4 cubed (*[tex \Large 4^3 = 64]) and 27 is 3 cubed (*[tex \Large 3^3 = 27]).


The factorization of the sum and difference of two cubes is:


*[tex \Large x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)]


This is really easy to remember:  1st factor: *[tex \Large ( x \text{  }\math y)]  2nd factor *[tex \Large (x^2 \text{  }\math xy \text{  }\math y^2)].  Then remember San Diego Padres = SDP = Same Different Plus to put in the signs.  So for  *[tex \Large x^3 + y^3 = (x + y)(x^2 - xy + y^2)] and for *[tex \Large x^3 - y^3 = (x - y)(x^2 + xy + y^2)]


So for your problem


*[tex \Large 64 + 27x^3 = 4^3 + (3x)^3 = (4 + 3x)(16 - 12x + 9x^2)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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