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16. \( 27 p^{6}-8 q^{12} \)

Ask by Chadwick Smith. in South Africa
Mar 08,2025

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\( 27p^6 - 8q^{12} = (3p^2 - 2q^4)(9p^4 + 6p^2q^4 + 4q^8) \).

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Beyond the Answer

Ah, the expression \( 27p^6 - 8q^{12} \) is a classic case of a difference of cubes! It can be represented as \( (3p^2)^3 - (2q^4)^3 \). You can factor it using the difference of cubes formula, which states that \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). In this case, we can let \( a = 3p^2 \) and \( b = 2q^4 \). So, when you factor it out, you'll end up with \( (3p^2 - 2q^4)((3p^2)^2 + (3p^2)(2q^4) + (2q^4)^2) \). This factored form can be helpful in many areas like algebra, helping to simplify complex expressions, or solving equations where such terms appear. When dealing with polynomial expressions, especially in higher level math, recognizing patterns like the difference of cubes can save you loads of time in factorization or finding roots!

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