Question
CISE 11 Factorise the followin (1) \( 27 x^{3}-1 \) (4) \( 125-729 x^{3} \)
Ask by Weber Salinas. in South Africa
Feb 12,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( 27x^{3} - 1 = (3x - 1)(9x^{2} + 3x + 1) \)
\( 125 - 729x^{3} = (5 - 9x)(25 + 45x + 81x^{2}) \)
Solution
We can factorise both expressions by recognising them as differences of cubes.
1) For 27x³ - 1:
Notice that 27x³ = (3x)³ and 1 = 1³. Using the formula for a difference of cubes, a³ - b³ = (a - b)(a² + ab + b²), let a = 3x and b = 1. Then:
27x³ - 1 = (3x - 1) [ (3x)² + (3x)(1) + 1² ]
= (3x - 1) (9x² + 3x + 1).
4) For 125 - 729x³:
Recognise that 125 = 5³ and 729x³ = (9x)³. Using the same formula with a = 5 and b = 9x, we have:
125 - 729x³ = (5 - 9x) [ 5² + 5(9x) + (9x)² ]
= (5 - 9x) (25 + 45x + 81x²).
Thus, the factorised forms are:
1) 27x³ - 1 = (3x - 1)(9x² + 3x + 1)
4) 125 - 729x³ = (5 - 9x)(25 + 45x + 81x²)
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Bonus Knowledge
Let's break down \( 27x^3 - 1 \) first! This expression is a difference of cubes since \( 27x^3 = (3x)^3 \) and \( 1 = 1^3 \). Using the difference of cubes factorization formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), we can rewrite it as \( (3x - 1)(9x^2 + 3x + 1) \). Now, onto \( 125 - 729x^3 \). This is also a difference of cubes, where \( 125 = 5^3 \) and \( 729x^3 = (9x)^3 \). Thus, we can again apply the same formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Therefore, we factor it as \( (5 - 9x)(25 + 45x + 81x^2) \). Happy factoring!
