Sure, let's go through the process of factoring each of these differences of two cubes step-by-step.
### 1. Factoring [tex]\( 8x^3 - 27y^3 \)[/tex]
The general formula for the difference of cubes is:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
For [tex]\( 8x^3 - 27y^3 \)[/tex]:
- Here, [tex]\( a = 2x \)[/tex] and [tex]\( b = 3y \)[/tex].
- Therefore:
[tex]\[
8x^3 - 27y^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) \\
= (2x - 3y)(4x^2 + 6xy + 9y^2)
\][/tex]
### 2. Factoring [tex]\( 16x^3 - 54y^3 \)[/tex]
First, factor out the common factor of 2:
[tex]\[ 16x^3 - 54y^3 = 2(8x^3 - 27y^3) \][/tex]
We have already factored [tex]\( 8x^3 - 27y^3 \)[/tex] from the first example:
[tex]\[ 8x^3 - 27y^3 = (2x - 3y)(4x^2 + 6xy + 9y^2) \][/tex]
Thus:
[tex]\[ 16x^3 - 54y^3 = 2(2x - 3y)(4x^2 + 6xy + 9y^2) \][/tex]
### 3. Factoring [tex]\( 27x^3 - 64y^3 \)[/tex]
Using the general formula again:
[tex]\[ a = 3x \text{ and } b = 4y \][/tex]
- Therefore:
[tex]\[
27x^3 - 64y^3 = (3x - 4y)((3x)^2 + (3x)(4y) + (4y)^2) \\
= (3x - 4y)(9x^2 + 12xy + 16y^2)
\][/tex]
### Matching each expression with its factored form:
[tex]\[ \begin{array}{ll}
8 x^3-27 y^3 & \rightarrow (2 x-3 y)\left(4 x^2+6 x y+9 y^2\right)\\
16 x^3-54 y^3 & \rightarrow 2(2 x-3 y)\left(4 x^2+6 x y+9 y^2\right)\\
27 x^3-64 y^3 & \rightarrow (3 x-4 y)\left(9 x^2+12 x y+16 y^2\right)\\
\end{array} \][/tex]
