Factor 64+q^(3) completely. Answer:

Recognize as sum of cubes: Recognize the expression as a sum of cubes. The given expression is 64 + q^3, which can be written as (4^3) + (q^3). This is a sum of cubes since 64 is 4 raised to the power of 3 and q^3 is q raised to the power of 3.Apply sum of cubes 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 = 4 and b = q.Substitute values into formula: Substitute the values of a and b into the formula. Using the values of a and b, we get: (4 + q)(4^2 - 4*q + q^2)Simplify terms in second factor: Simplify the terms in the second factor. Simplify 4^2 - 4*q + q^2 to get 16 - 4q + q^2.Write final factorized form: Write the final factorized form. The complete factorization of 64 + q^3 is: (4 + q)(16 - 4q + q^2)

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

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Algebra 1

Evaluate integers raised to positive rational exponents

Full solution

Q. Factor

64+q^{3}

completely.

\newline

Answer:

Recognize as sum of cubes: Recognize the expression as a sum of cubes. $\newline$ The given expression is $64 + q^3$ , which can be written as $(4^3) + (q^3)$ . $\newline$ This is a sum of cubes since $64$ is $4$ raised to the power of $3$ and $q^3$ is $q$ raised to the power of $3$ .
Apply sum of cubes formula: Apply the sum of cubes formula. $\newline$ The sum of cubes formula is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . $\newline$ Here, $a = 4$ and $b = q$ .
Substitute values into formula: Substitute the values of $a$ and $b$ into the formula. $\newline$ Using the values of $a$ and $b$ , we get: $\newline$ $(4 + q)(4^2 - 4\cdot q + q^2)$
Simplify terms in second factor: Simplify the terms in the second factor. $\newline$ Simplify $4^2 - 4q + q^2$ to get $16 - 4q + q^2$ .
Write final factorized form: Write the final factorized form. $\newline$ The complete factorization of $64 + q^3$ is: $\newline$ $(4 + q)(16 - 4q + q^2)$