Answer :
Sure, let's factor or simplify these polynomial expressions step-by-step:
### 42. Factor [tex]\( x^8 - 9 \)[/tex]
This expression resembles the difference of squares since [tex]\( 9 = 3^2 \)[/tex].
[tex]\[
x^8 - 9 = (x^4)^2 - 3^2 = (x^4 - 3)(x^4 + 3)
\][/tex]
### 43. Factor [tex]\( x^9 - 8 \)[/tex]
Notice that this can be expressed as a difference of cubes. [tex]\( 8 = 2^3 \)[/tex], so:
[tex]\[
x^9 - 8 = (x^3)^3 - 2^3 = (x^3 - 2)((x^3)^2 + 2x^3 + 4) = (x^3 - 2)(x^6 + 2x^3 + 4)
\][/tex]
### 44. Simplify [tex]\( 8x^3 + y^9 \)[/tex]
This expression does not factor further using simple identities, so it remains:
[tex]\[
8x^3 + y^9
\][/tex]
### 45. Factor [tex]\( x^6 - 27y^3 \)[/tex]
This can be expressed as a difference of cubes. [tex]\( 27 = 3^3 \)[/tex], so:
[tex]\[
x^6 - 27y^3 = (x^2)^3 - (3y)^3 = (x^2 - 3y)(x^4 + 3x^2y + 9y^2)
\][/tex]
### 46. Simplify [tex]\( 4x^2 + 144 \)[/tex]
This expression is a sum, not a standard form that we typically factor using identities like difference of squares, so it stays:
[tex]\[
4x^2 + 144
\][/tex]
### 47. Simplify [tex]\( 216 + 27y^{12} \)[/tex]
This expression can be factored by taking out common factors:
[tex]\[
216 + 27y^{12} = 27(8 + y^{12})
\][/tex]
Given in the context we wouldn't change further unless specific simplification is expected.
### 48. Simplify [tex]\( 64x^3 - 125y^6 \)[/tex]
This expression is a difference of cubes:
[tex]\[
64x^3 - 125y^6 = (4x)^3 - (5y^2)^3 = (4x - 5y^2)((4x)^2 + 4x(5y^2) + (5y^2)^2) = (4x - 5y^2)(16x^2 + 20xy^2 + 25y^4)
\][/tex]
### 49. Factor [tex]\(\frac{1}{16}x^6 - 25y^4\)[/tex]
This resembles a difference of squares:
[tex]\[
\frac{1}{16}x^6 - 25y^4 = \left(\frac{1}{4}x^3\right)^2 - (5y^2)^2 = \left(\frac{1}{4}x^3 - 5y^2\right)\left(\frac{1}{4}x^3 + 5y^2\right)
\][/tex]
Here you have the detailed process of factoring or simplifying each expression! If you have any more questions, feel free to ask.
### 42. Factor [tex]\( x^8 - 9 \)[/tex]
This expression resembles the difference of squares since [tex]\( 9 = 3^2 \)[/tex].
[tex]\[
x^8 - 9 = (x^4)^2 - 3^2 = (x^4 - 3)(x^4 + 3)
\][/tex]
### 43. Factor [tex]\( x^9 - 8 \)[/tex]
Notice that this can be expressed as a difference of cubes. [tex]\( 8 = 2^3 \)[/tex], so:
[tex]\[
x^9 - 8 = (x^3)^3 - 2^3 = (x^3 - 2)((x^3)^2 + 2x^3 + 4) = (x^3 - 2)(x^6 + 2x^3 + 4)
\][/tex]
### 44. Simplify [tex]\( 8x^3 + y^9 \)[/tex]
This expression does not factor further using simple identities, so it remains:
[tex]\[
8x^3 + y^9
\][/tex]
### 45. Factor [tex]\( x^6 - 27y^3 \)[/tex]
This can be expressed as a difference of cubes. [tex]\( 27 = 3^3 \)[/tex], so:
[tex]\[
x^6 - 27y^3 = (x^2)^3 - (3y)^3 = (x^2 - 3y)(x^4 + 3x^2y + 9y^2)
\][/tex]
### 46. Simplify [tex]\( 4x^2 + 144 \)[/tex]
This expression is a sum, not a standard form that we typically factor using identities like difference of squares, so it stays:
[tex]\[
4x^2 + 144
\][/tex]
### 47. Simplify [tex]\( 216 + 27y^{12} \)[/tex]
This expression can be factored by taking out common factors:
[tex]\[
216 + 27y^{12} = 27(8 + y^{12})
\][/tex]
Given in the context we wouldn't change further unless specific simplification is expected.
### 48. Simplify [tex]\( 64x^3 - 125y^6 \)[/tex]
This expression is a difference of cubes:
[tex]\[
64x^3 - 125y^6 = (4x)^3 - (5y^2)^3 = (4x - 5y^2)((4x)^2 + 4x(5y^2) + (5y^2)^2) = (4x - 5y^2)(16x^2 + 20xy^2 + 25y^4)
\][/tex]
### 49. Factor [tex]\(\frac{1}{16}x^6 - 25y^4\)[/tex]
This resembles a difference of squares:
[tex]\[
\frac{1}{16}x^6 - 25y^4 = \left(\frac{1}{4}x^3\right)^2 - (5y^2)^2 = \left(\frac{1}{4}x^3 - 5y^2\right)\left(\frac{1}{4}x^3 + 5y^2\right)
\][/tex]
Here you have the detailed process of factoring or simplifying each expression! If you have any more questions, feel free to ask.
