Complex algebra problem

[philosopher]

So here's the problem:



I've tried to solve it by first solving for Z. And what i get are 2 possible solutions which are (√3+i)/2, and (√3-i)/2, but when i plug them back into the equation to check it gets so messy and i don't get the initial solution. It would get even messier if i tried to plug it in the second one. There is probably easier method of solving this, right?

 

[Deleted member 4993]

So here's the problem:

View attachment 20059

I've tried to solve it by first solving for Z. And what i get are 2 possible solutions which are (√3+i)/2, and (√3-i)/2, but when i plug them back into the equation to check it gets so messy and i don't get the initial solution. It would get even messier if i tried to plug it in the second one. There is probably easier method of solving this, right?

Can you "factorize" the expression (a³ + b³)?

Please show us what you have tried and exactly where you are stuck.​

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Please follow the rules of posting in this forum, as enunciated at:​

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https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

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Please share your work/thoughts about this assignment.​

 

[pka]

So here's the problem:

View attachment 20059

I've tried to solve it by first solving for Z. And what i get are 2 possible solutions which are (√3+i)/2, and (√3-i)/2, but when i plug them back into the equation to check it gets so messy and i don't get the initial solution. It would get even messier if i tried to plug it in the second one. There is probably easier method of solving this, right?

We are given that \(\left(z+\dfrac{1}{z}\right)^2=3\) (note the square is positive real).
So I would say that \(\left(z+\dfrac{1}{z}\right)=\sqrt3\). Expand
\(\left(z+\dfrac{1}{z}\right)^3=z^3+\dfrac{1}{z^3}+3\left(z+\dfrac{1}{z}\right)\)
\(=z^3+\dfrac{1}{z^3}+3\sqrt3=~?\)
Can you finish?

 

[Dr.Peterson]

So here's the problem:

View attachment 20059

I've tried to solve it by first solving for Z. And what i get are 2 possible solutions which are (√3+i)/2, and (√3-i)/2, but when i plug them back into the equation to check it gets so messy and i don't get the initial solution. It would get even messier if i tried to plug it in the second one. There is probably easier method of solving this, right?

Please show your work, at least for the check, so we can see if you are making any mistakes. It shouldn't be that hard.

On the other hand, pka's suggestion is presumably the intended way to approach this problem. My concern with that kind of work is that occasionally it turns out that in fact there is no solution, so that the result is hypothetical only. In this case, you know there is a solution, so either way is safe, and his is easier (once you see it). Both are worth doing for the exercise!