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Factor
64+125a^(3) completely.
Answer:

Factor $64+125 a^{3}$ completely. $\newline$ Answer:

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Estimate positive square roots

Full solution

Q. Factor

64+125 a^{3}

completely.

\newline

Answer:

Recognize as sum of cubes: Recognize the expression as a sum of two cubes. The given expression is $64 + 125a^{3}$ . We can rewrite $64$ as $(4)^{3}$ and $125a^{3}$ as $(5a)^{3}$ . This allows us to express the given sum as a sum of cubes: $(4)^{3} + (5a)^{3}$ .
Apply sum of cubes formula: Apply the sum of cubes formula. $\newline$ The sum of cubes formula is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . We will apply this formula to our expression where $a = 4$ and $b = 5a$ .
Substitute values: Substitute the values into the sum of cubes formula. $\newline$ Using the values $a = 4$ and $b = 5a$ , we get the factorization $(4 + 5a)((4)^2 - (4)(5a) + (5a)^2)$ .
Simplify factors: Simplify the factors. $\newline$ Now we simplify each factor: $\newline$ First factor: $(4 + 5a)$ remains the same. $\newline$ Second factor: $(4)^2 - (4)(5a) + (5a)^2$ simplifies to $16 - 20a + 25a^2$ .
Write final factorization: Write the final factorization. $\newline$ The complete factorization of the expression is $(4 + 5a)(16 - 20a + 25a^2)$ .

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