Why it is different?

The results of raising things into power strictly depends on the writer's interpretation; whereas $\frac{6}{2}=3$, raising to $\frac{6}{2}$ is not the same as raising to $3$. First, let's focus on $\frac{6}{2}$.

This is because there's a difference between $\sqrt{(-2)^6}$ and $(\sqrt{-2})^6$.

(i) In one, you're ordering the operation raise this value into the sixth power and square root it

(ii) In another, you're ordering the operation square root this value and raise it into the sixth power.

Here's how they differ: In (i), you will be taking a square root of a single value, and will thus obtain two distinct roots (unless it's 0). In (ii), you will be finding the two distinct square roots, and then raising each of them individually into the sixth power. If your answer was in a complex form, or were being raised into an odd power, you would actually get two distinct answers upon raising into that power; but either way, the results shall differ.

Now, raising to $3$ is another story. When you choose to simplify $\frac{6}{2}$ to $3$, you're killing off roots. Take a look at the following example:

$x^2=x$.

Now, you may be tempted to cancel the $x$ on both sides and cheer $x=1$!, but then you'd be killing a root:

$x^2-x=0$

$x(x-1)=0$

$\boxed{x=0,1}$

(iii) Here, you're ordering the operation raise this value into the third power

In conclusion, it's just a matter of notation. This is one of those confusing wings of mathematics that teachers and professors leave out in high schools and colleges. And this is why generally mathematicians refer to "principal roots." But all of this jibber-jabr isn't for nothing. In real-world applications, depending on what you're trying to calculate, you may want to take the square root of a value to the sixth power to observe symmetry, or just cube it because your model is simplistic. It's all up to you.

Cheers,

The right one is :>

$(-2)^{\frac{6}{2}}$

$=(-2)^{3}$

$=-8$

But the question is why the other one is wrong.

I will try to explain it ::

$(-2)^{\frac{6}{2}}$

$=((-2)^{6})^{\frac{1}{2}}$ No doubt !!! This step is absolutely right.

$=\sqrt{(-2)^{6}}$

$=\sqrt{ (-2).(-2).(-2).(-2).(-2).(-2)}$

$=\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.\sqrt{-2}.$

$=\sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i \sqrt{2}.i$ (Here $i=\sqrt{-1}$ ,i.e. iota )

$=(\sqrt{2})^{6} . i^{6}$

$=8 \times -1$ (Because * $i^{6}=i^{4}.i^{2}=1 \times -1=-1)$ *

Hidden concept here :

$\sqrt{a} \times \sqrt{b}=\sqrt{ab}$ if and only if, at least one of a and b is positive

If both a and b are negative, then:

$\sqrt{a} \times \sqrt{b}=\sqrt{|a|}.i \times \sqrt{|b|}.i = i^{2}. \sqrt{|a|} \times \sqrt{|b|}=-\sqrt{|a|.|b|}$

Root of a square(or even power) always comes out with modulus

$\sqrt{x^{2}} = |x|$

These results are differents because ${ \left( -2 \right) }^{ \frac { 6 }{ 2 } }\neq { \left( { \left( -2 \right) }^{ 6 } \right) }^{ \frac { 1 }{ 2 } }$.

${ \left( { \left( -2 \right) }^{ 6 } \right) }^{ \frac { 1 }{ 2 } }=\sqrt { { \left( -2 \right) }^{ 6 } }$ ) not ${ \sqrt { \left( -2 \right) } }^{ 6 }$.

The correct way to solve this in the 2nd step in the pink box is ${ \left( { \left( -2 \right) }^{ \frac { 1 }{ 2 } } \right) }^{ 6 }$because ${ \left( { \left( -2 \right) }^{ \frac { 1 }{ 2 } } \right) }^{ 6 }={ \sqrt { \left( -2 \right) } }^{ 6 }={ \left( -2 \right) }^{ \frac { 6 }{ 2 } }$

when you square by odd the negative sign still stays and is possible. However, once you power by an even number, and the number will always be positive no matter what other number power you put it to after. Hence, the parenthesis should be around 6/2 because that should be done first, so the -2 isnt powered by an even exponent.

Another thing, when you take square root you should have two possible answers....8 and -8. b/c both squared gives you 64. So it works but the 8 is extraneous and you missed the -8

i dun get it , the pink one box :O

simply "BODMAS" first division then multiplication . In second case also we have divide first i.e., (1/2) then multiply...........

in (-2)^6/2 the square root after multiplying to power 6 may be +ve or -ve

You might've not concentrated enough. Look the right side of your question there you've written - 64 to the power 1/2 This is the same as square root of 64.

So at last you got your obvious answer as +/- 8. This is according to algebraic rules. Although you've not mentioned -8 there and written only +8.