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How do you factor #x^3-27#?

Answer 1

Nathan Axel

Use the difference of cubes identity to find:

#x^3-27 = (x-3)(x^2+3x+9)#

Both #x^3# and #27=3^3# are perfect cubes. So we can use the difference of cubes identity:

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

with #a=x# and #b=3# as follows:

#x^3-27#

#=x^3-3^3#

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

#=(x-3)(x^2+3x+9)#

This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:

#=(x-3)(x-3omega)(x-3omega^2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

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Answer 2

Jace Anderson

To factor ( x^3 - 27 ), you can use the difference of cubes formula. The formula states that ( a^3 - b^3 = (a - b)(a^2 + ab + b^2) ).

In this case, ( a = x ) and ( b = 3 ), because ( 27 = 3^3 ).

So, applying the difference of cubes formula:

( x^3 - 27 = (x - 3)(x^2 + 3x + 9) )

Therefore, ( x^3 - 27 ) factors into ( (x - 3)(x^2 + 3x + 9) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.