Which monomial is a perfect cube

Which monomial is a perfect cube? 16×6 27×8 32×12 64×6

The correct answer and explanation is:

Let’s analyze each monomial to determine which one is a perfect cube.

Given monomials:

1. 16x616x^6
2. 27x827x^8
3. 32x1232x^{12}
4. 64x664x^6

Step 1: Understand what makes a monomial a perfect cube

A perfect cube means the entire monomial can be expressed as (a⋅xb)3(a \cdot x^b)^3, where both the numerical coefficient aa and the variable term xbx^b are perfect cubes individually.

• The coefficient must be a perfect cube (e.g., 1,8,27,64,125,…1, 8, 27, 64, 125, \ldots)
• The exponent on the variable must be divisible by 3 because (xk)3=x3k(x^k)^3 = x^{3k}

Step 2: Check each monomial

1. 16x616x^6

• Coefficient: 16
  • 16 is not a perfect cube because 23=82^3 = 8, 33=273^3 = 27, so 16 is not a perfect cube.
• Exponent of xx: 6
  • 6 is divisible by 3, so x6=(x2)3x^6 = (x^2)^3 is a perfect cube term.

Since the coefficient is not a perfect cube, 16x^6 is not a perfect cube.

2. 27x827x^8

• Coefficient: 27
  • 27=3327 = 3^3, so this is a perfect cube.
• Exponent of xx: 8
  • 8 is not divisible by 3, so x8x^8 is not a perfect cube term.

Therefore, 27x^8 is not a perfect cube.

3. 32x1232x^{12}

• Coefficient: 32
  • 32=2532 = 2^5, which is not a perfect cube (since 23=82^3=8, 26=642^6=64)
• Exponent of xx: 12
  • 12 is divisible by 3, so x12=(x4)3x^{12} = (x^4)^3 is a perfect cube term.

Since the coefficient is not a perfect cube, 32x^{12} is not a perfect cube.

4. 64x664x^6

• Coefficient: 64
  • 64=4364 = 4^3 or 262^6, so 64 is a perfect cube.
• Exponent of xx: 6
  • 6 is divisible by 3, so x6=(x2)3x^6 = (x^2)^3 is a perfect cube.

Since both the coefficient and variable term are perfect cubes, 64x^6 is a perfect cube.

Final answer:

64×6\boxed{64x^6}

Explanation (about 300 words):

A monomial is a perfect cube if it can be written as some expression raised to the third power. For example, (axb)3=a3x3b(a x^b)^3 = a^3 x^{3b}. This means two conditions must be met:

1. The coefficient (numerical part) must itself be a perfect cube.
2. The exponent on the variable must be divisible by 3 because when you cube xbx^b, the exponent becomes 3b3b.

Let’s apply this to the four given monomials:

• 16x⁶: The coefficient 16 is not a perfect cube since 23=82^3 = 8 and 33=273^3 = 27, but 16 falls in between. Although the variable part x6x^6 can be expressed as (x2)3(x^2)^3, the coefficient prevents this monomial from being a perfect cube.
• 27x⁸: Here, 27 is a perfect cube (333^3), but the exponent on xx is 8, which is not divisible by 3, so it cannot be expressed as (xk)3(x^k)^3.
• 32x^{12}: The variable part is perfect cube because 12 is divisible by 3, but 32 is 252^5, not a perfect cube. So this monomial is not a perfect cube.
• 64x⁶: 64 is a perfect cube because 43=644^3 = 64. Also, 6 is divisible by 3, so x6=(x2)3x^6 = (x^2)^3. This monomial can be written as (4×2)3(4x^2)^3, confirming it is a perfect cube.

Therefore, the only monomial that is a perfect cube from the list is 64x^6.