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

Ask by Stewart Horton. in South Africa
Feb 03,2025

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Answer

Here are the factorizations for the given expressions: 1. \(27 x^{3}-1 = (3x-1)(9x^{2}+3x+1)\) 2. \(8 x^{3}+1 = (2x+1)(4x^{2}-2x+1)\) 3. \(64 x^{3}-y^{3} = (4x-y)(16x^{2}+4xy+y^{2})\) 4. \(125-729 x^{3} = (5-9x)(25+45x+81x^{2})\) 5. \(a^{3} b^{3}-\frac{1}{8} = \frac{1}{8}(2ab-1)(4a^{2}b^{2}+2ab+1)\) 6. \(5 x^{3}+40 = 5(x+2)(x^{2}-2x+4)\) 7. \(8 a^{4}-64 a = 8a(a-2)(a^{2}+2a+4)\) 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 more help with any of these, let me know!

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Extra Insights

First additional information: The expressions you see here are great examples of sum and difference of cubes! For instance, \(27x^3 - 1\) can be factorized using the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), where \(a = 3x\) and \(b = 1\). This gives you \((3x - 1)(9x^2 + 3x + 1)\). Second additional information: Enjoy the challenge of recognizing patterns in algebra! Mistakes can happen when we overlook the coefficients or forget to apply formulas consistently, especially in problems involving higher powers or complex binomials. Double-checking your substitutions and ensuring that all terms are factored correctly can save you from minor slip-ups!

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