Answer :
Certainly! Let's factorize each of the given expressions step by step.
### Part (a): \( 250 m^4 - 2 m \)
1. Identify the common factor:
Both terms have a factor of \( 2m \). So, we can factor out \( 2m \).
2. Factor out the common factor:
[tex]\[ 250 m^4 - 2 m = 2m (125 m^3 - 1) \][/tex]
3. Factor the remaining expression:
Notice that \( 125 m^3 - 1 \) is a difference of cubes, which can be factorized as:
[tex]\[ 125 m^3 = (5m)^3 \quad \text{and} \quad 1 = 1^3 \][/tex]
Thus,
[tex]\[ 125 m^3 - 1 = (5m - 1)(25m^2 + 5m + 1) \][/tex]
So, the fully factorized form is:
[tex]\[ 250 m^4 - 2 m = 2m (5m - 1) (25m^2 + 5m + 1) \][/tex]
### Part (b): \( x^3 y - 64 y^4 \)
1. Identify the common factor:
Both terms have a factor of \( y \). So, we can factor out \( y \).
2. Factor out the common factor:
[tex]\[ x^3 y - 64 y^4 = y (x^3 - 64 y^3) \][/tex]
3. Factor the remaining expression:
Notice that \( x^3 - 64 y^3 \) is a difference of cubes, which can be factorized as:
[tex]\[ 64 y^3 = (4y)^3 \][/tex]
Thus,
[tex]\[ x^3 - 64 y^3 = (x - 4y) (x^2 + 4xy + 16y^2) \][/tex]
So, the fully factorized form is:
[tex]\[ x^3 y - 64 y^4 = y (x - 4y) (x^2 + 4xy + 16y^2) \][/tex]
### Part (d): \( (a-b)^3 - 8(a+b)^3 \)
1. Recognize the cubes:
This can be recognized as a difference of cubes:
[tex]\[ 8(a+b)^3 = (2(a+b))^3 \][/tex]
2. Use the difference of cubes formula:
[tex]\[ (A)^3 - (B)^3 = (A - B)(A^2 + AB + B^2) \][/tex]
Here, \( A = a - b \) and \( B = 2(a + b) \):
Substituting, we get:
[tex]\[ (a-b)^3 - 8(a+b)^3 = \big((a - b) - 2(a + b)\big)\cdot\big((a-b)^2 + (a-b)\cdot2(a+b) + (2(a+b))^2\big) \][/tex]
3. Simplify the expression:
[tex]\[ (a - b) - 2(a + b) = a - b - 2a - 2b = -a - 3b \][/tex]
And,
[tex]\[ (a-b)^2 + 2(a-b)(a+b) + 4(a+b)^2 = a^2 - 2ab + b^2 + 2(a^2 - b^2) + 4a^2 + 8ab + 4b^2 = 7a^2 + 6ab + 3b^2 \][/tex]
So, the fully factorized form is:
[tex]\[ (a-b)^3 - 8(a+b)^3 = -(a + 3b)(7a^2 + 6ab + 3b^2) \][/tex]
### Part (e): \( \frac{x^3}{y^3} - \frac{y^3}{x^3} \)
1. Combine the terms under a common denominator:
[tex]\[ \frac{x^3}{y^3} - \frac{y^3}{x^3} = \frac{x^6 - y^6}{x^3 y^3} \][/tex]
2. Recognize the numerator as a difference of squares:
[tex]\[ x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \][/tex]
3. Factor further the differences of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
[tex]\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \][/tex]
Therefore, the full factorization becomes:
[tex]\[ \frac{x^6 - y^6}{x^3 y^3} = \frac{(x^3 - y^3)(x^3 + y^3)}{x^3 y^3} = \frac{(x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)}{x^3 y^3} \][/tex]
So, the fully factorized form is:
[tex]\[ \frac{x^3}{y^3} - \frac{y^3}{x^3} = \frac{(x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)}{x^3 y^3} \][/tex]
Now we have the fully factorized forms for each of the given expressions.
### Part (a): \( 250 m^4 - 2 m \)
1. Identify the common factor:
Both terms have a factor of \( 2m \). So, we can factor out \( 2m \).
2. Factor out the common factor:
[tex]\[ 250 m^4 - 2 m = 2m (125 m^3 - 1) \][/tex]
3. Factor the remaining expression:
Notice that \( 125 m^3 - 1 \) is a difference of cubes, which can be factorized as:
[tex]\[ 125 m^3 = (5m)^3 \quad \text{and} \quad 1 = 1^3 \][/tex]
Thus,
[tex]\[ 125 m^3 - 1 = (5m - 1)(25m^2 + 5m + 1) \][/tex]
So, the fully factorized form is:
[tex]\[ 250 m^4 - 2 m = 2m (5m - 1) (25m^2 + 5m + 1) \][/tex]
### Part (b): \( x^3 y - 64 y^4 \)
1. Identify the common factor:
Both terms have a factor of \( y \). So, we can factor out \( y \).
2. Factor out the common factor:
[tex]\[ x^3 y - 64 y^4 = y (x^3 - 64 y^3) \][/tex]
3. Factor the remaining expression:
Notice that \( x^3 - 64 y^3 \) is a difference of cubes, which can be factorized as:
[tex]\[ 64 y^3 = (4y)^3 \][/tex]
Thus,
[tex]\[ x^3 - 64 y^3 = (x - 4y) (x^2 + 4xy + 16y^2) \][/tex]
So, the fully factorized form is:
[tex]\[ x^3 y - 64 y^4 = y (x - 4y) (x^2 + 4xy + 16y^2) \][/tex]
### Part (d): \( (a-b)^3 - 8(a+b)^3 \)
1. Recognize the cubes:
This can be recognized as a difference of cubes:
[tex]\[ 8(a+b)^3 = (2(a+b))^3 \][/tex]
2. Use the difference of cubes formula:
[tex]\[ (A)^3 - (B)^3 = (A - B)(A^2 + AB + B^2) \][/tex]
Here, \( A = a - b \) and \( B = 2(a + b) \):
Substituting, we get:
[tex]\[ (a-b)^3 - 8(a+b)^3 = \big((a - b) - 2(a + b)\big)\cdot\big((a-b)^2 + (a-b)\cdot2(a+b) + (2(a+b))^2\big) \][/tex]
3. Simplify the expression:
[tex]\[ (a - b) - 2(a + b) = a - b - 2a - 2b = -a - 3b \][/tex]
And,
[tex]\[ (a-b)^2 + 2(a-b)(a+b) + 4(a+b)^2 = a^2 - 2ab + b^2 + 2(a^2 - b^2) + 4a^2 + 8ab + 4b^2 = 7a^2 + 6ab + 3b^2 \][/tex]
So, the fully factorized form is:
[tex]\[ (a-b)^3 - 8(a+b)^3 = -(a + 3b)(7a^2 + 6ab + 3b^2) \][/tex]
### Part (e): \( \frac{x^3}{y^3} - \frac{y^3}{x^3} \)
1. Combine the terms under a common denominator:
[tex]\[ \frac{x^3}{y^3} - \frac{y^3}{x^3} = \frac{x^6 - y^6}{x^3 y^3} \][/tex]
2. Recognize the numerator as a difference of squares:
[tex]\[ x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \][/tex]
3. Factor further the differences of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
[tex]\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \][/tex]
Therefore, the full factorization becomes:
[tex]\[ \frac{x^6 - y^6}{x^3 y^3} = \frac{(x^3 - y^3)(x^3 + y^3)}{x^3 y^3} = \frac{(x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)}{x^3 y^3} \][/tex]
So, the fully factorized form is:
[tex]\[ \frac{x^3}{y^3} - \frac{y^3}{x^3} = \frac{(x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)}{x^3 y^3} \][/tex]
Now we have the fully factorized forms for each of the given expressions.
