Melody

ok here is my logic.

\((x^2+\frac{1}{x})^n\)

let a be an integer such that    \(0\le a \le n\)

the a'th term will be

\(\begin{pmatrix}n\\a\end{pmatrix}[x^2]^a\;\left[ \frac{1}{x} \right]^{n-a}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{2a}\;\times x ^{a-n}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{3a-n}\\\ \)

this will be a constant term whenever

  \(3a-n=0\\n=3a\)

Since a is an integer, the expression will have a constant term if n is a multiple of 3

The value of the constant term is already included in my answer.

ok :)

If you know that the speed you travel in a car is

miles per hour    (if you are from the Uited States)    or

kilometers per hour  from most other places then you know that

Speed = distance per time

and 'per' means divide

so

speed = distance/time

Thanks Rom.

Thanks

Thanks for responding Cerenetie.

I am glad that you worked it out :) 

I hope my answer made you think a bit even though you went to your sister for more help :)

What calc did you put that into?

Yes that  is correct.

I think that you have already been given the answers to many of these. 

So it is time for you to start learning how to do them.

I will give you some examples that will help.

\(5^{1/2}=\sqrt5\\ 7^{-3}=\frac{1}{7^{+3}}\\ \\~\\ 8^{3/2}=(8^3)^{1/2}=(8^{1/2})^3\)

Now think about it and you tell me what  these are equal to.

\(1) \;\; 2^{-1/2}\\ 2) \;\; 2^{-3/2}\\ and \\ 3) \;\;b^{1/2}\)

Please no one answer over me

No it is not correct.    

I think i did it the hard way,....

(x-1)(x-2)(x-3)=(6-1)(6-2)(6-3)   

So x=6 is one solution.  

It is a cubic so there is a total of 3 answers for x.   

The other 2 satisfy   x^2+k=0

Maybe there is a quicker way to do it but I expanded the left.

\((x-1)(x-2)(x-3) =5*4*3\\ x^3-6x^2+11x-6=60\\ x^3-6x^2+11x-66=0\\ \text{Then I divided the LHS by (x-6) and got}\\ (x-6)(x^2+11)=0\\ So \\ x=6,\;and\;\;x^2=-11\\ x^2=-11\\ x=\pm\sqrt{-11}\\ x=x=\pm\sqrt{11}\;i\\ \)

But you were not asked for all the x solutions you were asked for k and  k=11