Why does #sqrt(x^6)# not equal #x^3#?

Question

Avery Alexander · Accepted Answer

See an explanation below:

#sqrt(x^6) = abs(x^3)# not #x^3# An example would be substituting #-2# for #x#: #sqrt(-2^6) = sqrt(64) = 8# #abs(-2^3) = abs(-8) = 8# #-2^3 = -8#

Jace Anderson · Answer

undefined The square root of x^6 does not equal x^3 because the square root of a number raised to an even power does not simplify to the number raised to half of that power. In this case, x^6 can be written as (x^2)^3, and taking the square root of x^6 would result in x^3, not x^2.

HIX Tutor · Answer

undefined That's a great question! Let's explore this step by step.

1. **What does sqrt mean?**  
   The square root (sqrt) of a number gives a value that, when multiplied by itself, equals that number.

2. **What is x^6?**  
   The expression x^6 means x multiplied by itself six times: x * x * x * x * x * x.

3. **What is the square root of x^6?**  
   When you find the square root of x^6, you want a number that, when multiplied by itself, gives x^6.

4. **How does it work with exponents?**  
   We know that:  
   $$
   \text{sqrt}(x^6) = x^{6/2} = x^3
   $$  
   This means that the square root of x^6 is indeed x^3, but there is more.

5. **What about negative numbers?**  
   If x is negative, like -2, what is the square root of (-2)^6?  
   (-2)^6 = 64, and √64 = 8.  
   But x would be -2, not 8. The square root gives us the positive answer.

6. **So, is sqrt(x^6) = x^3 for all x?**  
   Not always! It's true if x is non-negative, but if x is negative, the square root gives us a positive answer.

7. **Final thought:**  
   So, we say sqrt(x^6) gives us |x|^3, the absolute value of x cubed, because it ignores any negative sign.

Does that make sense? What do you think?