Which Monomial Is A Perfect Cube 1x10 8x8 9x9 27x15

Which Monomial is a Perfect Cube? A Step-by-Step Guide

Understanding the structure of algebraic expressions is fundamental to mastering higher mathematics. Among these expressions, monomials—single terms consisting of a coefficient and variables raised to non-negative integer exponents—hold a special place. So a key property to identify is whether a monomial is a perfect cube. This means the entire monomial can be expressed as some other monomial raised to the third power. Determining this requires a systematic check of both the numerical coefficient and the exponents of all variables. Given the options 1x¹⁰, 8x⁸, 9x⁹, and 27x¹⁵, only one satisfies the strict criteria for being a perfect cube. This article will demystify the process, providing you with a clear, repeatable method to solve such problems and build a stronger foundation in algebra.

Understanding the Criteria for a Perfect Cube Monomial

A monomial a * x^m * y^n * ... is a perfect cube if and only if two conditions are met simultaneously:

1. The coefficient a must itself be a perfect cube (i.Day to day, e. So , it can be written as k³ for some integer k). Think about it: 2. But Every exponent on the variables (m, n, ... But ) must be a multiple of 3 (i. In real terms, e. , divisible by 3 with no remainder).

This is because when you raise a monomial (k * x^p * y^q ...) to the third power, you get k³ * x^(3p) * y^(3q) .... Which means, to reverse-engineer this, the coefficient must be a cube, and each exponent must be 3 times some integer. This dual requirement is non-negotiable; failing either condition means the monomial is not a perfect cube.

Analyzing Each Option Systematically

Let’s apply this two-part test to each given monomial.

1. 1x¹⁰

• Coefficient Check: The coefficient is 1. Is 1 a perfect cube? Yes, because 1³ = 1. The coefficient passes.
• Exponent Check: The exponent on x is 10. Is 10 divisible by 3? 10 ÷ 3 = 3.333..., which is not an integer. So, 10 is not a multiple of 3.
• Verdict: Not a perfect cube. The variable exponent fails the test.

2. 8x⁸

• Coefficient Check: The coefficient is 8. Is 8 a perfect cube? Yes, because 2³ = 8. The coefficient passes.
• Exponent Check: The exponent on x is 8. Is 8 divisible by 3? 8 ÷ 3 ≈ 2.666..., not an integer. 8 is not a multiple of 3.
• Verdict: Not a perfect cube. The variable exponent fails the test.

3. 9x⁹

• Coefficient Check: The coefficient is 9. Is 9 a perfect cube? We need an integer k such that k³ = 9. 2³ = 8 and 3³ = 27. There is no integer between 2 and 3, so 9 is not a perfect cube.
• Exponent Check: The exponent on x is 9. 9 ÷ 3 = 3, which is an integer. So 9 is a multiple of 3. The exponent passes.
• Verdict: Not a perfect cube. Although the variable exponent is perfect, the coefficient 9 is not a perfect cube. This is a common trap. Both parts must be perfect cubes.

4. 27x¹⁵

• Coefficient Check: The coefficient is 27. Is 27 a perfect cube? Yes, because 3³ = 27. The coefficient passes.
• Exponent Check: The exponent on x is 15. 15 ÷ 3 = 5, which is an integer. So 15 is a multiple of 3. The exponent passes.
• Verdict: This is a perfect cube monomial. Both conditions are satisfied. We can write it as (3x⁵)³, since (3x⁵)³ = 3³ * (x⁵)³ = 27 * x¹⁵.

Common Pitfalls and How to Avoid Them

Students often make two critical errors when solving these problems. ** Seeing 8 or 27 might trigger an immediate "yes" without verifying the exponents. **First, they check only the coefficient.Remember, the entire term must be a cube.

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perfect cube, despite having a perfect cube coefficient, because the exponent 7 is not divisible by 3.

Second, they misapply the rule to negative coefficients or fractional exponents. For negative coefficients, remember that a negative number can be a perfect cube (e.g., (-2)³ = -8). So -8x⁹ is a perfect cube because -8 = (-2)³ and 9 is divisible by 3. That said, with fractional exponents, the numerator of the reduced fraction must be divisible by 3. Take this case: x^(2/3) is a perfect cube because (x^(2/9))³ = x^(2/3), but x^(4/3) is not because 4 is not divisible by 3.

Conclusion

Determining whether a monomial is a perfect cube hinges on a straightforward but strict two-part criterion: the numerical coefficient must be a perfect cube, and every variable exponent must be an integer multiple of 3. As demonstrated, overlooking either condition—whether by focusing solely on the coefficient, ignoring the divisibility of exponents, or mishandling signs and fractions—leads to incorrect classifications. Now, this test is both necessary and sufficient. Still, by systematically applying this dual check to each component of the monomial, you can confidently identify perfect cubes and avoid common traps. Mastery of this fundamental concept builds a crucial foundation for more advanced algebraic manipulation, including simplifying radical expressions and solving polynomial equations.

##Practice Problems

Applying the two‑part test to a variety of monomials helps solidify the concept. Work through each example, then check your answer against the brief explanation that follows.

Problem Set

1. ‑64x¹²y⁶

2. 125a⁹b⁴

3. ‑27x⁻⁹

4. 8x³⁄⁵y⁶

5. ‑1x⁰ ### Solutions

6. Coefficient: ‑64 = (‑4)³ → perfect cube.
   Exponents: 12 ÷ 3 = 4 (integer), 6 ÷ 3 = 2 (integer).
   Verdict: Perfect cube. It can be written as (‑4x⁴y²)³.

7. Coefficient: 125 = 5³ → perfect cube.
   Exponents: 9 ÷ 3 = 3 (integer) for a, but 4 ÷ 3 leaves a remainder.
   Verdict: Not a perfect cube because the exponent on b is not a multiple of 3.

8. Coefficient: ‑27 = (‑3)³ → perfect cube.
   Exponent: ‑9 ÷ 3 = ‑3 (integer; negative multiples are allowed).
   Verdict: Perfect cube. Expressed as (‑3x⁻³)³.

9. Coefficient: 8 = 2³ → perfect cube.
   Exponents: For x, the exponent is 3⁄5. Write it as a fraction 3⁄5; the numerator (3) is divisible by 3, so x³⁄⁵ = (x¹⁄⁵)³. For y, 6 ÷ 3 = 2 (integer).
   Verdict: Perfect cube. It equals (2x¹⁄⁵y²)³.

10. Coefficient: ‑1 = (‑1)³ → perfect cube.
    Exponent: 0 ÷ 3 = 0 (integer). Any variable to the zero power is 1, which trivially satisfies the condition.
    Verdict: Perfect cube. Indeed, (‑1)³ = ‑1.

Additional Tips for Mastery

• Factor first: If a monomial contains a common numerical factor, pull it out before testing. Here's a good example: 54x⁶ = 2·27x⁶; since 2 is not a cube, the whole term fails regardless of the exponent.
• Use prime factorization: Breaking the coefficient into primes makes it easy to see whether each prime’s exponent is a multiple of 3.
• Watch for hidden fractions: When an exponent is expressed as a decimal, convert it to a fraction in lowest terms before checking the numerator.
• Practice with negative bases: Remember that (‑a)³ = ‑a³, so a negative coefficient does not disqualify a monomial from being a perfect cube.

Conclusion

By consistently verifying that (1) the numerical coefficient is a perfect cube and (2) every variable exponent is an integer multiple of 3 (or, for fractional exponents, that the reduced numerator is divisible by 3), you can reliably classify any monomial as a perfect cube or not. The practice problems above illustrate how these rules apply to positive, negative, integer, and fractional cases. In practice, internalizing this two‑step check not only prevents common errors but also lays the groundwork for more advanced topics such as simplifying radical expressions, working with rational exponents, and solving higher‑degree polynomial equations. Keep practicing, and the distinction will become second nature.