Question 1:
To factor \( 27x^3 + 512y^3 \), we can recognize this expression as a sum of cubes. The formula for the sum of cubes \( a^3 + b^3 \) can be factored as:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case:
- \( a = 3x \) (since \( (3x)^3 = 27x^3 \))
- \( b = 8y \) (since \( (8y)^3 = 512y^3 \))
Using the sum of cubes formula, we have: \[ 27x^3 + 512y^3 = (3x + 8y)((3x)^2 - (3x)(8y) + (8y)^2) \] Calculating the second part: \[ = (3x + 8y)(9x^2 - 24xy + 64y^2) \]
From the options, one of the factors presented is \( 3x - 8y \) which does not match, and the cubic term \( 9x^2 - 24xy + 64y^2 \).
Thus, the correct answer is: 9x² - 24xy + 64y²
Question 2:
To factor \( x^{15} - 64 \), we recognize this as a difference of cubes. The formula for the difference of cubes \( a^3 - b^3 \) is given as:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, we can express \( x^{15} \) as \( (x^5)^3 \) and \( 64 \) as \( 4^3 \):
- \( a = x^5 \)
- \( b = 4 \)
Applying the difference of cubes formula: \[ x^{15} - 64 = (x^5 - 4)((x^5)^2 + (x^5)(4) + 4^2) \] Simplifying the second factor: \[ = (x^5 - 4)(x^{10} + 4x^5 + 16) \]
From the options, the first factor is \( x^5 - 4 \), which indeed is a factor.
Thus, the correct answer is: x5 - 4