ight)^3}{3} = - rac{2}{3} - rac{- rac{8}{27}}{3} = - rac{2}{3} + rac{8}{81} = - rac{54}{81} + rac{8}{81} = - rac{46}{81} $. Final answer: oxed{-\dfrac{46}{81}}

ight)^3}{3} = -rac{2}{3} - rac{-rac{8}{27}}{3} = -rac{2}{3} + rac{8}{81} = -rac{54}{81} + rac{8}{81} = -rac{46}{81} $. Final answer: oxed{-\dfrac{46}{81}}

["Understanding the Mathematical Expression: ight)³ ÷ 3 = –\frac{2}{3} – (–\frac{8}{27}) ÷ 3 = –\frac{2}{3} + \frac{8}{81} = –\frac{54}{81} + \frac{8}{81} = –\frac{46}{81", "Mathematics often involves intricate steps and elegant simplifications, and one such problem beautifully illustrates the power of careful calculation. Let’s break down the expression:\nRight)³ ÷ 3 = –\frac{2}{3} – (–\frac{8}{27}) ÷ 3 = –\frac{2}{3} + \frac{8}{81} = –\frac{54}{81} + \frac{8}{81} = –\frac{46}{81", "### Step-by-Step Breakdown", "1. Initial Expression\n Start with the equation:\n [\n \ ext{Right)}^3 \div 3 = -\frac{2}{3} - \left( -\frac{8}{27} \right) \div 3\n ]\n The double negative transforms into a simple multiplication:\n [\n -\frac{2}{3} + \frac{8}{27} \div 3\n ]", "2. Division Inside Parentheses\n Divide (-\frac{8}{27}) by 3:\n [\n -\frac{8}{27} ÷ 3 = -\frac{8}{27} \ imes \frac{1}{3} = -\frac{8}{81}\n ]\n Note: Dividing by a positive number flips the sign.", "3. Finding a Common Denominator\n To combine (-\frac{2}{3}) and (-\frac{8}{81}), express (-\frac{2}{3}) with denominator 81:\n [\n -\frac{2}{3} = -\frac{54}{81}\n ]", "4. Final Addition\n Now add:\n [\n -\frac{54}{81} + \frac{8}{81} = -\frac{46}{81}\n ]", "### Why This Simplification Matters", "This problem demonstrates essential algebra skills: manipulating fractions, applying the rule for division of signed numbers, and simplifying rational expressions. Mastering such steps builds confidence in solving more complex equations and prepares learners for advanced mathematical topics like calculus and algebra.", "### Final Answer\nThe simplified value of the given expression is:\n[\n\boxed{-\dfrac{46}{81}}\n]", "Understanding each operation step by step transforms a confusing equation into a clear, solvable form — proving that clarity in math begins with precision."]

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