Factor out the GCF of `4` from each term in the polynomial. `4(125x^3)+4(27)` Factor out the GCF of `4` from `500x^3+108` `4(125x^3+27)` Since both terms are perfect cubes, the binomial can be factored using the sum of cubesformula: `a^3+b^3 = (a+b)(a^2-ab+b^2)`

The answer will be, `4(5x+3)(25x^2-15x+9)`...

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Factor out the GCF of `4` from each term in the polynomial.

`4(125x^3)+4(27)`

Factor out the GCF of `4` from `500x^3+108`

`4(125x^3+27)`

Since both terms are perfect cubes, the binomial can be factored using the sum of cubes formula: `a^3+b^3 = (a+b)(a^2-ab+b^2)`

The answer will be,

`4(5x+3)(25x^2-15x+9)`

Factor `500x^3 + 108` .

First, factor common factor of 4.

`4(125x^3+27)`

Now, notice that `125x^3` is perfect cube because `5^3` = 125 and `x^3` is a cube. Also 27 is a cube as `3^3` = 27

Now apply the rule of the sum of cubes.

`a^3 + b^3 = (a + b)(a^2-2ab+b^2)`

This problem indicates `a = 5x` and `b = 3`

Therefore, `4(5x+3)(25x^2 -30x+9)`

**The factored solution to `500x^3+108` is `4(5x+3)(25x^2 -30x+9)` **

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