To solve the problem of factoring the binomial [tex]\(8x^3 + 125y^3\)[/tex], we need to use the sum of cubes formula. The sum of cubes formula is expressed as:
[tex]\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\][/tex]
First, we identify the terms [tex]\(a\)[/tex] and [tex]\(b\)[/tex] from the given binomial [tex]\(8x^3 + 125y^3\)[/tex]:
[tex]\[
a^3 = 8x^3 \quad \text{and} \quad b^3 = 125y^3
\][/tex]
Since [tex]\(8x^3 = (2x)^3\)[/tex] and [tex]\(125y^3 = (5y)^3\)[/tex], we can identify:
[tex]\[
a = 2x \quad \text{and} \quad b = 5y
\][/tex]
Next, applying the sum of cubes formula [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex]:
[tex]\[
8x^3 + 125y^3 = (2x + 5y)((2x)^2 - (2x)(5y) + (5y)^2)
\][/tex]
Now we expand and simplify the expression inside the parentheses:
[tex]\[
(2x + 5y)((2x)^2 - (2x)(5y) + (5y)^2)
\][/tex]
Calculate each term individually:
[tex]\[
(2x)^2 = 4x^2
\][/tex]
[tex]\[
(2x)(5y) = 10xy
\][/tex]
[tex]\[
(5y)^2 = 25y^2
\][/tex]
Putting it all together in the expression:
[tex]\[
(2x + 5y)(4x^2 - 10xy + 25y^2)
\][/tex]
From this, we can see that the missing term inside the parentheses is [tex]\(10xy\)[/tex].
Therefore, the missing term in the factorization [tex]\(8x^3 + 125y^3 = (2x + 5y)(4x^2 - ? + 25y^2)\)[/tex] is:
[tex]\[
10xy
\][/tex]
So, the answer is:
[tex]\[
\boxed{10xy}
\][/tex]
