Factor 125n^(3)-1 completely. Answer:

Recognize Cubes Expression: Recognize the expression as a difference of cubes. The expression 125n^3 - 1 can be written as (5n)^3 - 1^3, which is a difference of cubes since both 125 and 1 are perfect cubes.Apply Cubes Formula: Apply the difference of cubes formula. The difference of cubes formula is a^3 - b^3 = (a - b)(a^2 + ab + b^2). Here, a = 5n and b = 1.Substitute into Formula: Substitute a and b into the formula. Using the values of a and b from Step 2, we get: (5n)^3 - 1^3 = (5n - 1)((5n)^2 + (5n)(1) + 1^2)Expand and Simplify: Expand and simplify the terms in the formula. Now we expand (5n)^2, (5n)(1), and 1^2: (5n - 1)(25n^2 + 5n + 1)Check Final Factorization: Check the final expression for any further factorization. The quadratic 25n^2 + 5n + 1 does not factor further over the integers, so the factorization is complete.

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Factor $125 n^{3}-1$ completely. $\newline$ Answer:

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Algebra 2

Evaluate rational exponents

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Q. Factor

125 n^{3}-1

completely.

\newline

Answer:

Recognize Cubes Expression: Recognize the expression as a difference of cubes. The expression $125n^3 - 1$ can be written as $(5n)^3 - 1^3$ , which is a difference of cubes since both $125$ and $1$ are perfect cubes.
Apply Cubes Formula: Apply the difference of cubes formula. $\newline$ The difference of cubes formula is $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ . Here, $a = 5n$ and $b = 1$ .
Substitute into Formula: Substitute $a$ and $b$ into the formula. $\newline$ Using the values of $a$ and $b$ from Step $2$ , we get: $\newline$ $(5n)^3 - 1^3 = (5n - 1)((5n)^2 + (5n)(1) + 1^2)$
Expand and Simplify: Expand and simplify the terms in the formula. $\newline$ Now we expand $(5n)^2$ , $(5n)(1)$ , and $1^2$ : $\newline$ $(5n - 1)(25n^2 + 5n + 1)$
Check Final Factorization: Check the final expression for any further factorization. $\newline$ The quadratic $25n^2 + 5n + 1$ does not factor further over the integers, so the factorization is complete.