Answer :
Certainly! Let's work through the factorizations step by step for each expression.
1. Factorize [tex]\( 8 + a^3 \)[/tex]
To factor [tex]\( 8 + a^3 \)[/tex], we recognize this as a sum of cubes. The sum of cubes can be factored using the formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = a \)[/tex] and [tex]\( b = 2 \)[/tex] because [tex]\( 8 = 2^3 \)[/tex]. Applying the formula:
[tex]\[ 8 + a^3 = a^3 + 2^3 = (a + 2)(a^2 - 2a + 4) \][/tex]
So,
[tex]\[ 8 + a^3 = (a + 2)(a^2 - 2a + 4) \][/tex]
2. Factorize [tex]\( 27x^3 - 64 \)[/tex]
To factor [tex]\( 27x^3 - 64 \)[/tex], we recognize this as a difference of cubes. The difference of cubes can be factored using the formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = 3x \)[/tex] and [tex]\( b = 4 \)[/tex] because [tex]\( 27x^3 = (3x)^3 \)[/tex] and [tex]\( 64 = 4^3 \)[/tex]. Applying the formula:
[tex]\[ 27x^3 - 64 = (3x)^3 - 4^3 = (3x - 4)( (3x)^2 + (3x)(4) + 4^2) \][/tex]
Simplify the second factor:
[tex]\[ (3x)^2 + 3x \cdot 4 + 4^2 = 9x^2 + 12x + 16 \][/tex]
So,
[tex]\[ 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) \][/tex]
3. Factorize [tex]\( 125 + v^6 \)[/tex]
To factor [tex]\( 125 + v^6 \)[/tex], we recognize this as a sum of squares and cubes. This expression can be rewritten as:
[tex]\[ 125 + v^6 = (5)^3 + (v^2)^3 \][/tex]
Using the sum of cubes formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = v^2 \)[/tex] and [tex]\( b = 5 \)[/tex]. Applying the formula:
[tex]\[ v^6 + 125 = (v^2 + 5)(v^4 - 5v^2 + 25) \][/tex]
So,
[tex]\[ 125 + v^6 = (v^2 + 5)(v^4 - 5v^2 + 25) \][/tex]
4. Factorize [tex]\( 1000 - y^3 \)[/tex]
To factor [tex]\( 1000 - y^3 \)[/tex], we recognize this as a difference of cubes. Using the difference of cubes formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = 10 \)[/tex] and [tex]\( b = y \)[/tex] because [tex]\( 1000 = 10^3 \)[/tex]. Applying the formula:
[tex]\[ 1000 - y^3 = 10^3 - y^3 = (10 - y)((10)^2 + 10y + y^2) \][/tex]
[tex]\[ = (10 - y)(100 + 10y + y^2) \][/tex]
To make the leading term positive, we can rewrite it as:
[tex]\[ = -(y - 10)(y^2 + 10y + 100) \][/tex]
So,
[tex]\[ 1000 - y^3 = -(y - 10)(y^2 + 10y + 100) \][/tex]
5. Factorize [tex]\( 1 + m^3 n^3 \)[/tex]
To factor [tex]\( 1 + m^3 n^3 \)[/tex], we recognize this as a sum of cubes. Using the sum of cubes formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = mn \)[/tex] and [tex]\( b = 1 \)[/tex]. Applying the formula:
[tex]\[ 1 + m^3 n^3 = (mn)^3 + 1^3 = (mn + 1)((mn)^2 - mn \cdot 1 + 1^2) \][/tex]
[tex]\[ = (mn + 1)(m^2 n^2 - mn + 1) \][/tex]
So,
[tex]\[ 1 + m^3 n^3 = (mn + 1)(m^2 n^2 - mn + 1) \][/tex]
Therefore, the factorizations are:
1. [tex]\( 8 + a^3 = (a + 2)(a^2 - 2a + 4) \)[/tex]
2. [tex]\( 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) \)[/tex]
3. [tex]\( 125 + v^6 = (v^2 + 5)(v^4 - 5v^2 + 25) \)[/tex]
4. [tex]\( 1000 - y^3 = -(y - 10)(y^2 + 10y + 100) \)[/tex]
5. [tex]\( 1 + m^3 n^3 = (mn + 1)(m^2 n^2 - mn + 1) \)[/tex]
1. Factorize [tex]\( 8 + a^3 \)[/tex]
To factor [tex]\( 8 + a^3 \)[/tex], we recognize this as a sum of cubes. The sum of cubes can be factored using the formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = a \)[/tex] and [tex]\( b = 2 \)[/tex] because [tex]\( 8 = 2^3 \)[/tex]. Applying the formula:
[tex]\[ 8 + a^3 = a^3 + 2^3 = (a + 2)(a^2 - 2a + 4) \][/tex]
So,
[tex]\[ 8 + a^3 = (a + 2)(a^2 - 2a + 4) \][/tex]
2. Factorize [tex]\( 27x^3 - 64 \)[/tex]
To factor [tex]\( 27x^3 - 64 \)[/tex], we recognize this as a difference of cubes. The difference of cubes can be factored using the formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = 3x \)[/tex] and [tex]\( b = 4 \)[/tex] because [tex]\( 27x^3 = (3x)^3 \)[/tex] and [tex]\( 64 = 4^3 \)[/tex]. Applying the formula:
[tex]\[ 27x^3 - 64 = (3x)^3 - 4^3 = (3x - 4)( (3x)^2 + (3x)(4) + 4^2) \][/tex]
Simplify the second factor:
[tex]\[ (3x)^2 + 3x \cdot 4 + 4^2 = 9x^2 + 12x + 16 \][/tex]
So,
[tex]\[ 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) \][/tex]
3. Factorize [tex]\( 125 + v^6 \)[/tex]
To factor [tex]\( 125 + v^6 \)[/tex], we recognize this as a sum of squares and cubes. This expression can be rewritten as:
[tex]\[ 125 + v^6 = (5)^3 + (v^2)^3 \][/tex]
Using the sum of cubes formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = v^2 \)[/tex] and [tex]\( b = 5 \)[/tex]. Applying the formula:
[tex]\[ v^6 + 125 = (v^2 + 5)(v^4 - 5v^2 + 25) \][/tex]
So,
[tex]\[ 125 + v^6 = (v^2 + 5)(v^4 - 5v^2 + 25) \][/tex]
4. Factorize [tex]\( 1000 - y^3 \)[/tex]
To factor [tex]\( 1000 - y^3 \)[/tex], we recognize this as a difference of cubes. Using the difference of cubes formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = 10 \)[/tex] and [tex]\( b = y \)[/tex] because [tex]\( 1000 = 10^3 \)[/tex]. Applying the formula:
[tex]\[ 1000 - y^3 = 10^3 - y^3 = (10 - y)((10)^2 + 10y + y^2) \][/tex]
[tex]\[ = (10 - y)(100 + 10y + y^2) \][/tex]
To make the leading term positive, we can rewrite it as:
[tex]\[ = -(y - 10)(y^2 + 10y + 100) \][/tex]
So,
[tex]\[ 1000 - y^3 = -(y - 10)(y^2 + 10y + 100) \][/tex]
5. Factorize [tex]\( 1 + m^3 n^3 \)[/tex]
To factor [tex]\( 1 + m^3 n^3 \)[/tex], we recognize this as a sum of cubes. Using the sum of cubes formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Here, [tex]\( a = mn \)[/tex] and [tex]\( b = 1 \)[/tex]. Applying the formula:
[tex]\[ 1 + m^3 n^3 = (mn)^3 + 1^3 = (mn + 1)((mn)^2 - mn \cdot 1 + 1^2) \][/tex]
[tex]\[ = (mn + 1)(m^2 n^2 - mn + 1) \][/tex]
So,
[tex]\[ 1 + m^3 n^3 = (mn + 1)(m^2 n^2 - mn + 1) \][/tex]
Therefore, the factorizations are:
1. [tex]\( 8 + a^3 = (a + 2)(a^2 - 2a + 4) \)[/tex]
2. [tex]\( 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) \)[/tex]
3. [tex]\( 125 + v^6 = (v^2 + 5)(v^4 - 5v^2 + 25) \)[/tex]
4. [tex]\( 1000 - y^3 = -(y - 10)(y^2 + 10y + 100) \)[/tex]
5. [tex]\( 1 + m^3 n^3 = (mn + 1)(m^2 n^2 - mn + 1) \)[/tex]
