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I am given $\left(x+iy\right)^{\frac{1}{3}}=a+ib$, and I need to prove $4\left(a^{2\ }-b^{2}\right)=\frac{x}{a}+\frac{y}{b}$.

The first "key" thing (I hope it's actually useful!) I notice is that we are only considering the real parts (I'm pretty sure, anyway).

I tried "cheating" but extending the LHS and RHS to $4\left(a-b\right)\left(a+b\right)=\frac{xb+ya}{ab}$ but I don't think that's particularly helpful for me because I don't immediately know what to do.

I also tried rewriting the given pieces of information: $\left(a+ib\right)^{3}$ to find

$x=a^{3}-3ab^{2}$

$y=3a^{2}b-b^{3}$

Right now my plan in to get everything on LHS then RHS in terms of $x$ and $y$ to show they are equal, but I am having trouble with that. If you have a better method as well, feel free to comment.

José Carlos Santos
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user71207
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  • $$\frac xa +\frac yb$$ just putting $x=a(a^2-3b^2),y=b(3a^2-b^2)$. – lone student Apr 01 '21 at 13:13
  • Note, there is a tacit assumption here that should really be made explicit, namely that $ab\not=0$, so that you're not dividing by $0$ in the expression ${x\over a}+{y\over b}$. – Barry Cipra Apr 01 '21 at 13:19

1 Answers1

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You already (correctly) figured out that $x+iy = (a+ib)^3$ implies $$ x = a(a^2-3b^2) \, ,\\ y = b(3a^2 -b^2) \, . $$ It follows that $$ \frac x a + \frac y b = (a^2-3b^2)+ (3a^2 -b^2) = 4(a^2 -b^2) $$

Another way is to compute $$ (x+iy)(a+ib) = (a+ib)^4 = (a^4-6a^2b^2+b^4) + i(4a^3 b-4ab^3) $$ and compare the imaginary parts: $$ xb + ya = 4a^3 b-4ab^3 = 4ab (a^2-b^2) \, . $$ If $ab \ne 0$ then the desired formula follows.

Martin R
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  • Wow, glad my work was actually relevant for once! – user71207 Apr 01 '21 at 13:11
  • How would you go from LHS to RHS though? Also will the "everything in terms of y and x" method work? I suspect no because its very hard to find a and b in terms of x and y – user71207 Apr 01 '21 at 13:12
  • @user71207: A non-zero complex number has three distinct cube roots, therefore you cannot uniquely determine $a$ and $b$ from $x$ and $y$. – Martin R Apr 01 '21 at 13:14
  • Nice for method 2. So it seems going from RHS to LHS is the way to go. Then LHS to RHS is not simply done? – user71207 Apr 02 '21 at 01:25