Answer :
Certainly! Let's analyze each binomial to determine whether it can be factored using either the difference of squares or the difference of cubes. We'll consider the necessary forms for each type of factorization.
1. Binomial: [tex]\( 64a^3 - b^6 \)[/tex]
- Difference of Cubes: To factor a binomial as a difference of cubes, it needs to be of the form [tex]\( A^3 - B^3 \)[/tex].
- Here, [tex]\( 64a^3 \)[/tex] can be written as [tex]\( (4a)^3 \)[/tex] and [tex]\( b^6 \)[/tex] can be written as [tex]\( (b^2)^3 \)[/tex].
- Hence, [tex]\( 64a^3 - b^6 \)[/tex] can be written as [tex]\( (4a)^3 - (b^2)^3 \)[/tex], making it a difference of cubes.
Conclusion: This binomial can be factored using the Difference of Cubes.
2. Binomial: [tex]\( 25x^8 - 9y^2 \)[/tex]
- Difference of Squares: To factor a binomial as a difference of squares, it needs to be of the form [tex]\( A^2 - B^2 \)[/tex].
- Here, [tex]\( 25x^8 \)[/tex] can be written as [tex]\( (5x^4)^2 \)[/tex] and [tex]\( 9y^2 \)[/tex] can be written as [tex]\( (3y)^2 \)[/tex].
- Hence, [tex]\( 25x^8 - 9y^2 \)[/tex] can be written as [tex]\( (5x^4)^2 - (3y)^2 \)[/tex], making it a difference of squares.
Conclusion: This binomial can be factored using the Difference of Squares.
3. Binomial: [tex]\( n^4 - 16 \)[/tex]
- Difference of Squares: To factor a binomial as a difference of squares, it needs to be of the form [tex]\( A^2 - B^2 \)[/tex].
- Here, [tex]\( n^4 \)[/tex] can be written as [tex]\( (n^2)^2 \)[/tex] and [tex]\( 16 \)[/tex] can be written as [tex]\( 4^2 \)[/tex].
- Hence, [tex]\( n^4 - 16 \)[/tex] can be written as [tex]\( (n^2)^2 - 4^2 \)[/tex], making it a difference of squares.
Conclusion: This binomial can be factored using the Difference of Squares.
4. Binomial: [tex]\( 16x^4y - 2xy^4 \)[/tex]
- Simplifying: First, we factor out the common factor [tex]\( 2xy \)[/tex].
- [tex]\( 16x^4y - 2xy^4 = 2xy(8x^3 - y^3) \)[/tex].
- Difference of Cubes: To factor a binomial as a difference of cubes, it needs to be of the form [tex]\( A^3 - B^3 \)[/tex].
- Here, [tex]\( 8x^3 \)[/tex] can be written as [tex]\( (2x)^3 \)[/tex] and [tex]\( y^3 \)[/tex] remains [tex]\( y^3 \)[/tex].
- Hence, [tex]\( 8x^3 - y^3 \)[/tex] can be written as [tex]\( (2x)^3 - y^3 \)[/tex], making it a difference of cubes.
Conclusion: This binomial can be factored using the Difference of Cubes (after factoring out the common factor).
Putting it all together:
- [tex]\( 64a^3 - b^6 \)[/tex]: Difference of Cubes
- [tex]\( 25x^8 - 9y^2 \)[/tex]: Difference of Squares
- [tex]\( n^4 - 16 \)[/tex]: Difference of Squares
- [tex]\( 16x^4y - 2xy^4 \)[/tex]: Difference of Cubes
So, the answer is:
- [tex]\( [1, 0, 0, 0] \)[/tex]
In summary:
- [tex]\( 64a^3 - b^6 \)[/tex] -> Difference of Cubes
- [tex]\( 25x^8 - 9y^2 \)[/tex] -> Difference of Squares
- [tex]\( n^4 - 16 \)[/tex] -> Difference of Squares
- [tex]\( 16x^4y - 2xy^4 \)[/tex] -> Difference of Cubes
1. Binomial: [tex]\( 64a^3 - b^6 \)[/tex]
- Difference of Cubes: To factor a binomial as a difference of cubes, it needs to be of the form [tex]\( A^3 - B^3 \)[/tex].
- Here, [tex]\( 64a^3 \)[/tex] can be written as [tex]\( (4a)^3 \)[/tex] and [tex]\( b^6 \)[/tex] can be written as [tex]\( (b^2)^3 \)[/tex].
- Hence, [tex]\( 64a^3 - b^6 \)[/tex] can be written as [tex]\( (4a)^3 - (b^2)^3 \)[/tex], making it a difference of cubes.
Conclusion: This binomial can be factored using the Difference of Cubes.
2. Binomial: [tex]\( 25x^8 - 9y^2 \)[/tex]
- Difference of Squares: To factor a binomial as a difference of squares, it needs to be of the form [tex]\( A^2 - B^2 \)[/tex].
- Here, [tex]\( 25x^8 \)[/tex] can be written as [tex]\( (5x^4)^2 \)[/tex] and [tex]\( 9y^2 \)[/tex] can be written as [tex]\( (3y)^2 \)[/tex].
- Hence, [tex]\( 25x^8 - 9y^2 \)[/tex] can be written as [tex]\( (5x^4)^2 - (3y)^2 \)[/tex], making it a difference of squares.
Conclusion: This binomial can be factored using the Difference of Squares.
3. Binomial: [tex]\( n^4 - 16 \)[/tex]
- Difference of Squares: To factor a binomial as a difference of squares, it needs to be of the form [tex]\( A^2 - B^2 \)[/tex].
- Here, [tex]\( n^4 \)[/tex] can be written as [tex]\( (n^2)^2 \)[/tex] and [tex]\( 16 \)[/tex] can be written as [tex]\( 4^2 \)[/tex].
- Hence, [tex]\( n^4 - 16 \)[/tex] can be written as [tex]\( (n^2)^2 - 4^2 \)[/tex], making it a difference of squares.
Conclusion: This binomial can be factored using the Difference of Squares.
4. Binomial: [tex]\( 16x^4y - 2xy^4 \)[/tex]
- Simplifying: First, we factor out the common factor [tex]\( 2xy \)[/tex].
- [tex]\( 16x^4y - 2xy^4 = 2xy(8x^3 - y^3) \)[/tex].
- Difference of Cubes: To factor a binomial as a difference of cubes, it needs to be of the form [tex]\( A^3 - B^3 \)[/tex].
- Here, [tex]\( 8x^3 \)[/tex] can be written as [tex]\( (2x)^3 \)[/tex] and [tex]\( y^3 \)[/tex] remains [tex]\( y^3 \)[/tex].
- Hence, [tex]\( 8x^3 - y^3 \)[/tex] can be written as [tex]\( (2x)^3 - y^3 \)[/tex], making it a difference of cubes.
Conclusion: This binomial can be factored using the Difference of Cubes (after factoring out the common factor).
Putting it all together:
- [tex]\( 64a^3 - b^6 \)[/tex]: Difference of Cubes
- [tex]\( 25x^8 - 9y^2 \)[/tex]: Difference of Squares
- [tex]\( n^4 - 16 \)[/tex]: Difference of Squares
- [tex]\( 16x^4y - 2xy^4 \)[/tex]: Difference of Cubes
So, the answer is:
- [tex]\( [1, 0, 0, 0] \)[/tex]
In summary:
- [tex]\( 64a^3 - b^6 \)[/tex] -> Difference of Cubes
- [tex]\( 25x^8 - 9y^2 \)[/tex] -> Difference of Squares
- [tex]\( n^4 - 16 \)[/tex] -> Difference of Squares
- [tex]\( 16x^4y - 2xy^4 \)[/tex] -> Difference of Cubes
