How do you factor #x^3+8#?

1 Answer
Mar 14, 2018

Answer:

Read below.

Explanation:

An interesting fact:

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

In #x^3+8#, #a^3=x^3# and #b^3=8#

Let's solve for #a# and #b#.

#=>a^3=x^3#

#=>root [3] (a^3)= root[3] (x^3)#

#=>a= x#

Now for #b#.

#=>b^3=8#

#=>root [3] (b^3)= root[3] (8)#

#=>b= 2#

Plug these values into our equation.

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

#(x+2)(x^2-2x+4)# This is our answer!

If you want to factor this further, we let #x^2-2x+4=0# and solve the equation.

#x^2-2x+4=0# Use the quadratic formula:
#(-b+-sqrt(b^2-4(a)(c)))/(2(a))#

Here, #a=1#, #b=-2#, and #c=4#

#x=(-(-2)+-sqrt((-2)^2-4(1)(4)))/(2(1))#

#x=(2+-sqrt(4-16))/(2)#

#x=(2+-sqrt(-12))/(2)#

#x=(2+-2isqrt(3))/(2)#

#x=1+-isqrt(3)#

#(x+2)(x-(1+isqrt3))(x-(1-isqrt3))#

#(x+2)(x-1-isqrt3))(x-1+isqrt3))#

Therefore, our factored form, in this case, would be
#(x+2)(x-1-isqrt3))(x-1+isqrt3))#