Factor 27-125v^(3) completely. Answer:

Identify Type of Factoring: Identify the type of factoring required for 27 - 125v^3. This is a difference of cubes since both terms are perfect cubes: 27 = 3^3 and 125v^3 = (5v)^3.Write Down Formula: Write down the formula for factoring a difference of cubes. The formula for factoring a difference of cubes a^3 - b^3 is (a - b)(a^2 + ab + b^2).Apply Formula to 27 - 125v^3: Apply the formula to 27 - 125v^3. Let a = 3 and b = 5v, then we have: (3 - 5v)(3^2 + 3*5v + (5v)^2)Calculate Squares and Product: Calculate the squares and the product of a and b. (3 - 5v)(9 + 15v + 25v^2)Write Final Factored Form: Write the final factored form. The factored form of 27 - 125v^3 is (3 - 5v)(9 + 15v + 25v^2).

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Math Problems

Algebra 1

Factor quadratics: special cases

Full solution

Q. Factor

27-125 v^{3}

completely.

\newline

Answer:

Identify Type of Factoring: Identify the type of factoring required for $27 - 125v^3$ . This is a difference of cubes since both terms are perfect cubes: $27 = 3^3$ and $125v^3 = (5v)^3$ .
Write Down Formula: Write down the formula for factoring a difference of cubes. $\newline$ The formula for factoring a difference of cubes $a^3 - b^3$ is $(a - b)(a^2 + ab + b^2)$ .
Apply Formula to $27 - 125v^3$ : Apply the formula to $27 - 125v^3$ . $\newline$ Let $a = 3$ and $b = 5v$ , then we have: $\newline$ $(3 - 5v)(3^2 + 3\cdot 5v + (5v)^2)$
Calculate Squares and Product: Calculate the squares and the product of $a$ and $b$ . $(3 - 5v)(9 + 15v + 25v^2)$
Write Final Factored Form: Write the final factored form. $\newline$ The factored form of $27 - 125v^3$ is $(3 - 5v)(9 + 15v + 25v^2)$ .