Factor 27n^(3)+64 completely. Answer:

Recognize Cubes: Recognize the expression as a sum of two cubes. The expression 27n^3 + 64 can be written as (3n)^3 + 4^3, which is a sum of cubes since 27n^3 is the cube of 3n and 64 is the cube of 4.Apply Formula: Apply the sum of cubes formula. The sum of cubes formula is a^3 + b^3 = (a + b)(a^2 - ab + b^2). Here, a = 3n and b = 4.Substitute a and b: Substitute a and b into the sum of cubes formula. Using the formula from Step 2, we substitute a = 3n and b = 4 to get: (3n + 4)((3n)^2 - (3n)(4) + 4^2)Simplify Expression: Simplify the expression. Now we simplify each part of the expression: (3n + 4)(9n^2 - 12n + 16)Check Factorization: Check for any further factorization. The quadratic expression 9n^2 - 12n + 16 does not factor further since it has no real roots (the discriminant b^2 - 4ac is negative).

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Factor $27 n^{3}+64$ completely. $\newline$ Answer:

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Grade 7

Powers with negative bases

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Q. Factor

27 n^{3}+64

completely.

\newline

Answer:

Recognize Cubes: Recognize the expression as a sum of two cubes. The expression $27n^3 + 64$ can be written as $(3n)^3 + 4^3$ , which is a sum of cubes since $27n^3$ is the cube of $3n$ and $64$ is the cube of $4$ .
Apply Formula: Apply the sum of cubes formula. $\newline$ The sum of cubes formula is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . Here, $a = 3n$ and $b = 4$ .
Substitute $a$ and $b$ : Substitute $a$ and $b$ into the sum of cubes formula. $\newline$ Using the formula from Step $2$ , we substitute $a = 3n$ and $b = 4$ to get: $\newline$ $(3n + 4)((3n)^2 - (3n)(4) + 4^2)$
Simplify Expression: Simplify the expression. $\newline$ Now we simplify each part of the expression: $\newline$ $(3n + 4)(9n^2 - 12n + 16)$
Check Factorization: Check for any further factorization. $\newline$ The quadratic expression $9n^2 - 12n + 16$ does not factor further since it has no real roots (the discriminant $b^2 - 4ac$ is negative).