Factorize: 8x^6 + 5x^3 + 1(use a^3 + b^3 + c^3 -3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac) (^-raised to)

Hello student,
Please find the answer to your question below
we checking rational solutions find that there is no rational root and hence there is no term of the form (x+a)
now 8x^6 = (2x^2)^3 and 1= 1^3 if we can express 5x^3 so that one is a cube and second one is product of -x 2 and the 3rd term then we get into the form a^3+b^3+c^3-3abc
8x^6+5x^3+1 = 8x^6+(-x)^3 + 1^3 +6x^3
= 8x^6 - x^3 + 1^3 + 3(x^2)(x)
= a^3+b^3+c^3 - 3abc where a = 2x^2 , b = -x and c = +1
hence it factors as (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (2x^2 -x +1)(4x^4+x^2+1 + 2x^3 +2x -2x^2)
=(2x^2-x+1)(4x^4+2x^3-x^2+2x+1)