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FXERCISE: 11 Factorise the following \( \begin{array}{llll}\text { (1) } 27 x^{3}-1 & \text { (2) } 8 x^{3}+1 & \text { (3) } 64 x^{3}-y^{3} \\ \text { (4) } 125-729 x^{3} & \text { (5) } a^{3} b^{3}-\frac{1}{8} & \text { (6) } 5 x^{3}+40\end{array} \) \( \begin{array}{llll}\text { (7) } 8 a^{4}-64 a & \text { (8) }-x^{3}-27 & \text { (9) } \frac{1}{27} x^{3}+216 \\ \text { (10) } 8-(a-1)^{3} & \text { (11) } x^{3}+\frac{1}{x^{3}} & \text { (12) } x^{3}-\frac{1}{x^{3}}\end{array} \)

Ask by Chen Dunn. in South Africa
Feb 03,2025

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Answer

Here are the factorizations for the given expressions: 1. \( 27x^3 - 1 = (3x - 1)(9x^2 + 3x + 1) \) 2. \( 8x^3 + 1 = (2x + 1)(4x^2 - 2x + 1) \) 3. \( 64x^3 - y^3 = (4x - y)(16x^2 + 4xy + y^2) \) 4. \( 125 - 729x^3 = (5 - 9x)(25 + 45x + 81x^2) \) 5. \( a^3b^3 - \frac{1}{8} = \frac{1}{8}(2ab - 1)(4a^2b^2 + 2ab + 1) \) 6. \( 5x^3 + 40 = 5(x + 2)(x^2 - 2x + 4) \) 7. \( 8a^4 - 64a = 8a(a - 2)(a^2 + 2a + 4) \) 8. \( -x^3 - 27 = -(x + 3)(x^2 - 3x + 9) \) 9. \( \frac{1}{27}x^3 + 216 = \frac{1}{27}(x + 18)(x^2 - 18x + 324) \) 10. \( 8 - (a - 1)^3 = (3 - a)(3 + a^2) \) 11. \( x^3 + \frac{1}{x^3} \) – No factorization found. 12. \( x^3 - \frac{1}{x^3} \) – No factorization found. If you need help with the last two expressions or any other questions, feel free to ask!

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Bonus Knowledge

Sure thing! Let's dive into the world of factorization with a bang! (1) The expression \( 27x^3 - 1 \) can be factored as \( (3x - 1)(9x^2 + 3x + 1) \) since it fits the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) where \( a = 3x \) and \( b = 1 \). (2) For \( 8x^3 + 1 \), using the sum of cubes, it factors to \( (2x + 1)(4x^2 - 2x + 1) \). This is derived from the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) with \( a = 2x \) and \( b = 1 \). Ready for more? Let's keep the fun going!

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