Factorise each of the following:
(i) 27y 3 + 125z 3
(ii) 64m3 – 343n 3
[ Hint : See Question 9. ]
- Sum of cubes: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
- Difference of cubes: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
Step 1: Recognize as sum of cubes:
\( 27y^3 = (3y)^3 \), \( 125z^3 = (5z)^3 \)
Step 2: Apply sum of cubes formula \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \)
Here, \( a = 3y \), \( b = 5z \)
\[ 27y^3 + 125z^3 = (3y + 5z) \big((3y)^2 - (3y)(5z) + (5z)^2\big) \]
\[ = (3y + 5z)(9y^2 - 15yz + 25z^2) \]
Step 1: Recognize as difference of cubes:
\( 64m^3 = (4m)^3 \), \( 343n^3 = (7n)^3 \)
Step 2: Apply difference of cubes formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)
Here, \( a = 4m \), \( b = 7n \)
\[ 64m^3 - 343n^3 = (4m - 7n) \big((4m)^2 + (4m)(7n) + (7n)^2\big) \]
\[ = (4m - 7n)(16m^2 + 28mn + 49n^2) \]
If \( x = \left( 2 + \sqrt{3} \right)^3 + \left( 2 - \sqrt{3} \right)^{-3} \) and \( x^3 - 3x + k = 0 \), then the value of \( k \) is: