Please help me with another test question. It is super hard!

[suckatmath3]

I am having trouble with the first thinking question. I do not know what to do. I think the teacher said I have to only use one side. Can anyone please help me? I try doing it but I got it wrong.

 

[BigBeachBanana]

Which question do you need help with?

 

[BigBeachBanana]

 

[Dr.Peterson]

It's a little fuzzy, but I seem to be reading it correctly:

I started with what is to be shown and worked back to the given; then you can reverse the process. I think it's easier to see the connection this way.

 

[mmm4444bot]

Which question do you need help with?

The first question under the label "THINKING".

 

[MaxWong]

There was cubes on the left hand side of the given equation, yet no cube in the equation that needs to be proved. We try to work backwards so that there is cubes in the equation:
$$\begin{array}{rcl} \log \left(\dfrac{a - b}2\right) &=& \dfrac13 (\log a + 2 \log b)\\ 3 \log \left(\dfrac{a - b}2\right) &=& \log a + 2 \log b\\ \log \left(\left(\dfrac{a - b}2\right)^3\right) &=& \log a + 2 \log b\\ \log \left((a - b)^3\right) - 3 \log 2 &=& \log a + 2 \log b \end{array}$$
If we can prove that $\log \left((a - b)^3\right) - 3 \log 2 = \log a + 2 \log b$, then it is done.

We were given $a^3 - b^3 = \cdots$. The equation involves $(a - b)^3$. This hints us to use the identity $a^3 - b^3 = (a - b)^3 + 3ab(a - b)$. Can you continue from here?