Factor completely: $8 a ^ { 3 } + 1$
A) $( 2 a + 1 ) ( 4 a^ 2 - 2 a + 1 )$
B) $( 2 a + 1 ) ( 2 a - 1 ) ^ { 2 }$
C) $( 2 a + 1 ) ^ { 3 }$
D) $( 2 a - 1 ) ( 4 a ^2 + 2 a + 1 )$

Question

Factor completely: $8 a ^ { 3 } + 1$
A) $( 2 a + 1 ) ( 4 a^ 2 - 2 a + 1 )$ 
B) $( 2 a + 1 ) ( 2 a - 1 ) ^ { 2 }$
C) $( 2 a + 1 ) ^ { 3 }$ 
D) $( 2 a - 1 ) ( 4 a ^2 + 2 a + 1 )$

Quizplus · Accepted Answer

This expression is a sum of cubes, factored as $(2a)^3 + 1^3$, which follows the formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Applying this formula, we get $(2a + 1)(4a^2 - 2a + 1)$.

Factor Completely 8a3+18 a ^ { 3 } + 18a3+1 A) (2a+1)(4a2−2a+1)( 2 a + 1 ) ( 4 a^ 2 - 2 a + 1 )(2a+1)(4a2−2a+1)

Factor Completely $8 a ^ { 3 } + 1$ A) $( 2 a + 1 ) ( 4 a^ 2 - 2 a + 1 )$