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Question
factoring diff / sum of cubes (i
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question
factor ( n^3 + 1 ) completely.
answer attempt 1 out of 2
Step1: Identify sum of cubes
$n^3 + 1 = n^3 + 1^3$
Step2: Apply sum of cubes formula
Use $a^3+b^3=(a+b)(a^2-ab+b^2)$, where $a=n$, $b=1$
$n^3 + 1^3=(n+1)(n^2 - n\cdot1 + 1^2)$
Step3: Simplify the quadratic term
$(n+1)(n^2 - n + 1)$
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$(n+1)(n^2 - n + 1)$