How do you factor #x^3 - 81#?

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Answer 1 · 2015-10-10T11:35:06

Express as a difference of cubes, then use the difference of cubes identity to find:

#x^3-81 = (x-3^(4/3))(x^2+3^(4/3)x+3^(8/3))#

When #a >= 0# and #b, c != 0#

#a^(bc) = (a^b)^c#

So:

#81 = 3^4 = 3^(4/3 * 3) = (3^(4/3))^3#

The difference of cubes identity is:

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

So:

#x^3 - 81 = x^3 - (3^(4/3))^3#

#=(x-3^(4/3))(x^2 + x(3^(4/3)) + (3^(4/3))^2)#

#=(x-3^(4/3))(x^2+3^(4/3)x+3^(8/3))#