Question

Here's a tricky question I saw somewhere, and I've shown it to my friends to see if the know to deal with it, (But I'm not sure if my version of solving it is correct or not ! )

I wan't to know where's the problem :(

We know that \[\LARGE (-3)^3=-27\]

but...

\[\LARGE (-3)^3=\left[(-3)^2\right]^{\frac32}=\left(3^2\right)^{\frac32}=27\]

and we know that:
\[\LARGE (-3)^3\neq27\]

where's the problem?

(if you wan't I can show you how I think it should be)- with square roots !

@Saifoo.khan @FoolForMath @ash2326 @Mertsj ...

Answer 1

sorry for disturbing :( ...

Answer 2

Math killer question.

Answer 3

Would you like me to show what I've done ? @saifoo.khan

Answer 4

You do got the correct point though. But im still confused why we are getting different answers.

Answer 5

When you have
\[\huge {(({-3})^2})^{\frac{3}{2}}\]
this is actually
\[\huge (-3)^\frac{6}{2}\]
which is
\[\huge ({\sqrt {-3}})^6\]
I think square root has to be evaluated first

Answer 6

@ash2326 that's what I'm thinking too ... :(

Answer 7

Ah! Ash got a point though.

I think it's similar to the radical equations we have in which after finding the solutions we have to check the answers.

Answer 8

@Kreshnik are using latex to change color of your posts????

Answer 9

Nope @ash2326 , tags change the colour.

Answer 10

what color ? O_O

Answer 11

@Kreshnik , this highlighting thing.

Answer 12

@saifoo.khan Ohk:)

Answer 13

It didn't change for me @saifoo.khan

Answer 14

You @ash2326 and @saifoo.khan have posts with something like yellow color , Because you mentioned me, as I did to you ! that's why is highlighted !

Answer 15

\[(-3)^{3}=[(-3)^{2}]^{3\div 2}\]....this can't be written as because in this case it will give you imaginary value...because of in the square root never will be a negative numaber.. and therefore you cant break this in this way...

Answer 16

\[\LARGE (\sqrt[]{-3})^3=i(\sqrt{-3})^3 \quad \quad \text{??}\]

if \[\LARGE i=\sqrt{-1}\]

is

\[\huge i=(\sqrt[x]{-1})^y\]

??

Answer 17

Yeah @Taufique is correct, we can't have a negative term inside a square root

Answer 18

Uhh.. sorry, \[\LARGE (\sqrt{-3})^3=i(\sqrt{3})^3\] this is what I wanted to ask , I made a mistake there :(

Answer 19

Rule of indices:

\[\Huge2^{2^{2^{2}}} = 2^{2^ \left({2^{2} } \right) }= 2^{16} =65536\]

But,

\[\Huge2^{2^{2^{2}}} = \left(2^2 \right) ^{ \left(2^2 \right)} = 256 \]

Answer 20

Only the First is correct by the current rule of indices.

Answer 21

Thanks everyone for helping... ;)

Answer 22

Btw I don't think the negative number thing is cogent here. This is simply a problem of using wrong rule of indices.

Answer 23

\[\sqrt{-3} ^{3}\]\[=\sqrt{-1\times3} ^{3}\]\[=\sqrt{-1}^3 \times\sqrt{3} ^{3}\]\[=i^3 \times\sqrt{3} ^{3}\]\[=-i \times3\sqrt{3}\], I think...

Answer 24

we can write X^n=(X^2)^n/2....when n is even number it is true for all X.
other wise when n is odd then X must be non negative number..

Answer 25

Rational powers m/n, where m/n is in lowest terms, are positive if m is even, negative for negative b if m and n are odd, and can be either sign if b is positive and n is even. (−27)1/3 = −3, (−27)2/3 = 9, and 43/2 has two roots 8 and −8. Since there is no real number x such that x2 = −1, the definition of bm/n when b is negative and n is even must use the imaginary unit i, as described more fully in the section Powers of complex numbers.
http://en.wikipedia.org/wiki/Exponentiation

Answer 26

EDIT: \[ \Huge2^{2^{2^{2}}} \neq \left(2^2 \right) ^{ \left(2^2 \right)} \neq 256 \]

Answer 27

Since there is no real number x such that \(x^2 = −1\), the definition of \(\huge b^{\frac{m}{n}}\) when b is negative and n is even must use the imaginary unit i, as described more fully in the section Powers of complex numbers.

Answer 28

Thanks again to everyone for the help :)