Exponents

Using the fractional exponent rule, write 27 as 3^3: 27 = 3^3. Therefore, (27)^{2/3} = (3^3)^{2/3} = 3^{(3×2/3)} = 3^2 = 9. Hence, option A is correct. An efficient exam method is to take the cube root first and then square: ∛27 = 3, so 3^2 = 9. Option 3 gives only the cube root, while 27 is the original number.

Write 64 as a perfect cube: \(64=4^3\). Therefore, \((64)^{1/3}=(4^3)^{1/3}=4\), so option A is correct. Option 8 is incorrect because \(8^3=512\), not 64. Exam tip: an exponent of \(1/n\) represents the nth root of a number.

When a power is raised to another power, the exponents are multiplied: \((a^m)^n=a^{m\times n}=a^{mn}\). Therefore, option A is correct. Option B is incorrect because adding exponents applies to multiplication of powers with the same base, not to a power raised to another power. Exam tip: for “power of a power,” multiply the exponents.

Using the laws of exponents, division of like bases gives \(x^m/x^n=x^{m-n}\). Multiplication of like bases then adds the exponents, so \(x^{m-n}\cdot x^p=x^{m-n+p}\). Option C incorrectly adds the exponents in the quotient, while option D uses the wrong sign for \(p\). This simplification is valid for \(x\neq0\). Exam tip: subtract exponents when dividing like bases and add them when multiplying like bases.

Since 8 = 2^3, x = 3, and since 27 = 3^3, y = 3. Therefore, x + y = 3 + 3 = 6, so option C is correct. In exponent questions, first determine each variable separately and then perform the required operation.

Using exponent rules we get (x+y=12) and (x-y=2), solving gives (x=7, y=5). Form equations and solve for quick exam accuracy.

The governing concept is the inverse relationship between cubing and taking a cube root. Let the positive number be x. Then x³ = 27. Since 27 = 3 × 3 × 3 = 3³, taking the positive cube root gives x = 3. The distractors do not work: 9³ = 729, while 27³ and 81³ are much larger than 27. Therefore, option A is correct.