Answer :
To solve the problem of factoring the binomial [tex]\(8x^3 + 27y^3\)[/tex], we can use the Sum of Cubes formula. The Sum of Cubes formula states that [tex]\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)[/tex].
Given [tex]\(8x^3 + 27y^3\)[/tex], we can identify the terms as follows:
- [tex]\(8x^3\)[/tex] is equivalent to [tex]\((2x)^3\)[/tex]
- [tex]\(27y^3\)[/tex] is equivalent to [tex]\((3y)^3\)[/tex]
So, we can rewrite the binomial as:
[tex]\[ (2x)^3 + (3y)^3 \][/tex]
According to the Sum of Cubes formula, this can be factored into:
[tex]\[ (2x + 3y)\left((2x)^2 - (2x)(3y) + (3y)^2\right) \][/tex]
Next, we will expand the terms inside the parentheses on the right-hand side:
1. [tex]\((2x)^2 = 4x^2\)[/tex]
2. [tex]\( (2x)(3y) = 6xy \)[/tex]
3. [tex]\((3y)^2 = 9y^2\)[/tex]
Therefore, the factored form is:
[tex]\[ (2x + 3y)(4x^2 - 6xy + 9y^2) \][/tex]
The missing term in the expression [tex]\(4x^2 - ? + 9y^2\)[/tex] is [tex]\(6xy\)[/tex]. Thus, without the minus sign, the missing term is:
[tex]\[ \boxed{xyxyxyxyxyxy} \][/tex]
Given [tex]\(8x^3 + 27y^3\)[/tex], we can identify the terms as follows:
- [tex]\(8x^3\)[/tex] is equivalent to [tex]\((2x)^3\)[/tex]
- [tex]\(27y^3\)[/tex] is equivalent to [tex]\((3y)^3\)[/tex]
So, we can rewrite the binomial as:
[tex]\[ (2x)^3 + (3y)^3 \][/tex]
According to the Sum of Cubes formula, this can be factored into:
[tex]\[ (2x + 3y)\left((2x)^2 - (2x)(3y) + (3y)^2\right) \][/tex]
Next, we will expand the terms inside the parentheses on the right-hand side:
1. [tex]\((2x)^2 = 4x^2\)[/tex]
2. [tex]\( (2x)(3y) = 6xy \)[/tex]
3. [tex]\((3y)^2 = 9y^2\)[/tex]
Therefore, the factored form is:
[tex]\[ (2x + 3y)(4x^2 - 6xy + 9y^2) \][/tex]
The missing term in the expression [tex]\(4x^2 - ? + 9y^2\)[/tex] is [tex]\(6xy\)[/tex]. Thus, without the minus sign, the missing term is:
[tex]\[ \boxed{xyxyxyxyxyxy} \][/tex]
