Find the cube - roots of :
`-216xx1728`

Answer

To find the cube roots of the expression \(-216 \times 1728\), we can follow these steps:

### Step 1: Calculate the product
Let \( A = -216 \times 1728 \).

### Step 2: Factor the numbers
We need to find the prime factors of both numbers.

**Factors of 216:**
- \( 216 = 2 \times 108 \)
- \( 108 = 2 \times 54 \)
- \( 54 = 2 \times 27 \)
- \( 27 = 3 \times 9 \)
- \( 9 = 3 \times 3 \)

So, the prime factorization of \( 216 \) is:
\[ 216 = 2^3 \times 3^3 \]

**Factors of 1728:**
- \( 1728 = 2 \times 864 \)
- \( 864 = 2 \times 432 \)
- \( 432 = 2 \times 216 \)
- \( 216 = 2^3 \times 3^3 \) (as calculated above)

So, we can continue factoring \( 1728 \):
- \( 1728 = 2^6 \times 3^3 \)

### Step 3: Combine the factors
Now, we can combine the factors of \( -216 \) and \( 1728 \):
\[ A = - (2^3 \times 3^3) \times (2^6 \times 3^3) \]

### Step 4: Simplify the expression
Combine the powers of the same bases:
\[ A = - (2^{3+6} \times 3^{3+3}) = - (2^9 \times 3^6) \]

### Step 5: Find the cube root
Now, we will find the cube root of \( A \):
\[ A^{1/3} = (- (2^9 \times 3^6))^{1/3} \]

Using the property of cube roots:
\[ A^{1/3} = - (2^{9/3} \times 3^{6/3}) = - (2^3 \times 3^2) \]

### Step 6: Calculate the final value
Now calculate \( 2^3 \) and \( 3^2 \):
\[ 2^3 = 8 \quad \text{and} \quad 3^2 = 9 \]
Thus,
\[ A^{1/3} = - (8 \times 9) = -72 \]

### Final Answer
The cube root of \(-216 \times 1728\) is:
\[ \boxed{-72} \]

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