How do you factor completely #a^3 - 8#?

Question

Savannah Anderson · Accepted Answer

#a^3-8 = (a-2)(a^2+2a+4)#
#=(a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)#

Use the difference of cubes identity, which can be written: #A^3-B^3=(A-B)(A^2+AB+B^2)# Note that both #a^3# and #8=2^3# are both perfect cubes. So we find: #a^3-8# #=a^3-2^3# #=(a-2)(a^2+(a)(2)+2^2)# #=(a-2)(a^2+2a+4)# The remaining quadratic factor has negative discriminant, so no Real zeros and no linear factors with Real coefficients. It is possible to factor it using Complex coefficients: #a^2+2a+4# #=a^2+2a+1+3# #=(a+1)^2+3# #=(a+1)^2-(sqrt(3)i)^2# #= (a+1-sqrt(3)i)(a+1+sqrt(3)i)# So: #a^8-8 = (a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)# Another way to express the full factoring is: #a^3-8 = (a-2)(a-2omega)(a-2omega^2)# where #omega = -1/2+sqrt(3)/2 i# is the primitive Complex cube root of #1#.

Eva Abramson · Answer

undefined To factor completely $a^3 - 8$, we can use the difference of cubes formula, which states that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

In this case, we have $a^3 - 8$, which can be rewritten as $a^3 - 2^3$, where $b = 2$.

Now, applying the difference of cubes formula:

$$a^3 - 8 = (a - 2)(a^2 + 2a + 4)$$

So, $a^3 - 8$ factors completely into $(a - 2)(a^2 + 2a + 4)$.