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))#
