Factors of 8a³ + b³ − 6ab + 1
The factors of \[ 8a^3+b^3-6ab+1 \] are
(a) \((2a+b-1)(4a^2+b^2+1-3ab-2a)\)
(b) \((2a-b+1)(4a^2+b^2-4ab+1-2a+b)\)
(c) \((2a+b+1)(4a^2+b^2+1-2ab-b-2a)\)
(d) \((2a-1+b)(4a^2+1-4a-b-2ab)\)
Solution
\[ (2a)^3+b^3+1^3-3(2a)(b)(1) \]
Using identity:
\[ x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \]
\[ =(2a+b+1)(4a^2+b^2+1-2ab-b-2a) \]
Therefore,
\[ \boxed{(c)\ (2a+b+1)(4a^2+b^2+1-2ab-b-2a)} \]