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# A Simple, Yet Difficult, Math Problem

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You can't solve it like a normal algebraic system.
Or, to put it a bit more precisely, a and b are complex, but not real, numbers. Algebra remains the same, but requires a new term, $i$, such that $i^2 = -1$.

Cool problem. Took me an embarrassing amount of time to get it right, but fortunately, I had a long, worthless meeting and a palm pilot, so looked like I was just keeping careful meeting notes. ;)

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Actually there is an elegant solution to this using just algebra without any complex numbers. Hint, a^3+b^3 is part of (a+:doh:^3, ;)

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While we're on the topic of simple problems, I'll throw one in...

a + b = -3 and ab = 4, then a^3 + b^3 equal...

enjoy!

a^3+b^3=9

thanks :hihi:

Actually there is an elegant solution to this using just algebra without any complex numbers. Hint, a^3+b^3 is part of (a+;)^3, ;)

Ok, let me see.

The first post states that (a + B)=(-3). Max then says that (a^3+b^3)=9.

(a+B)^3=(a^3+3a^2b+3ab^2+b^3)

= (a^3+b^3)+3(a^2b+ab^2)

= (a^3+b^3)+3(4a+4b) [substituting ab=4]

= (a^3+b^3)+12(a+B)

= (a^3+b^3)+12(-3) [substituting (a+B)=-3

= (a^3+b^3)-36

= (-3)^3 [using original value of (a+B)=(-3)]

=(-27)

(a^3+b^3)-36=-27

(a^3+b^3)=9

This much I can gather, and accept. However, now attempt to solve it for a and b.

This is where Craig chimed in suggesting that a and b are not real but complex numbers.

This much I had not thought about, expecting it to be a simple algebraic equation of whole numbers.

a + b = -3 and ab = 4 suggesting that the 4 needed to be negative, which would then allow me to suggest an answer other than 9. Of course, now having proven that Max is correct with an answer of 9 let's step backward and solve the ab.

a=4/b

(a^3+b^3)=9

[64/(b^3) + b^3] = 9

64/(b^3) = 9 - (b^3)

64 = 9b^3 - 1

65/9 = (b^3)

Then using Max's value of 9.

(a^3) = 9 - (65/9)

a = [(81-65)/9]^(1/3)=(1.77777777777777)^(1/3)

b = (65/9)^(1/3)= (7.222222222222)^(1/3)

Sorry I don't have my calculator on me.

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Hint, a^3+b^3 is part of (a+b)^3
Neat, actually. For any sum and product, given any integer, one can find an expression using those for lower exponents.

So, if a + b = s and ab = p, what's:

$a^n + b^n$

in terms of s, p and n?

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The first post states that (a + b)=(-3).
It also, importantly, gives $ab = 4$
However, now attempt to solve it for a and b.

This is where Craig chimed in suggesting that a and b are not real but complex numbers.

Here’s my blunt, inelegant algebra, which lead to complex solutions for a and b that confirm Nootropic and Max’s result:

Given

$a + b = -3$

$ab = 4$

Then

$a = \frac4b$

$\frac4b +b +3 =0$

$b^2 +3b +4 =0$

$b = \frac{-3 \pm \sqrt{9-16}}2 = -\frac32 \pm \frac{\sqrt7}2 i$

So a solution is

$b = -\frac32 + \frac{\sqrt7}2 i$

$a = -3 –b = -3 +\frac32 - \frac{\sqrt7}2 i = -\frac32 - \frac{\sqrt7}2 i$

Substituting (to avoid all those ugly numeric constant):

$c=-\frac32$, $d= \frac{\sqrt7}2$

$a = c +di$, $b= c -di$

Then

$a^3 +b^3$

$= c^3 +3c^2di -3cd^2 -d^3i$

$+c^3 -3c^2di -3cd^2 +d^3i$

$= 2c^3 -6cd^2$

Substuting for the constants back,

$a^3 +b^3 = \frac{-54 +6 \cdot 3 \cdot 7}8 = 9$

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Well congratulations to Craig for proving me wrong (hooray for algebra!). However, there is much, much simpler way to solve it...

a^3+b^3 = (a+:evil:^3 - (3a^2b + 3ab^2) ...relying on the binomial theorem

factor out 3ab from the middle term...

= (a+:D^3 - 3ab(a+B) ...subsituting in ab = 4 and a+b = -3

= (-3)^3 - 3(4)(-3)

= - 27 - -36 = 9 ta da!

This seems a likely corollary from the binomial theorem, but I'm not sure if it would have to be proven separately that (I'm doubting a nautral number counterexample, but still):

a^n + b^n = (a+B)^n - (n(a^n-1*B)...+...(n(a*b^n-1))

so if (a+B) = s and ab = p...

(s)^n - (p(a^n-2...+...b^n-2)

However, this is where it becomes tricky as you cannot simply factor out an n, and as I was browsing through pascal's triangle, I noticed that some rows of the triangle would not factor cleanly into nice, natural numbers, of course this excludes one, since that's the coefficient of what we're interested in, so in row six, the coefficients would be 6, 15, 20, 15, 6, of which there is no natural factor (though I suppose if you really wanted a numerical answer, it would be no problem), so you can't really say you could "factor out an n", as it is so conveniently done from the case where n = 3 (if you recall your binomial theorem, where the coefficients are given by combinations, given by n!/(r!(n-r)!, the best that can be done is to factor out an "ab", since that would be greatest common factor between all the terms. Hopefully someone has a more creative insight here...

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It also, importantly, gives $ab = 4$Here’s my blunt, inelegant algebra, which lead to complex solutions for a and b that confirm Nootropic and Max’s result:

Given

$a + b = -3$

$ab = 4$

Then

$a = \frac4b$

$\frac4b +b +3 =0$

$b^2 +3b +4 =0$

$b = \frac{-3 \pm \sqrt{9-16}}2 = -\frac32 \pm \frac{\sqrt7}2 i$

So a solution is

$b = -\frac32 + \frac{\sqrt7}2 i$

$a = -3 –b = -3 +\frac32 - \frac{\sqrt7}2 i = -\frac32 - \frac{\sqrt7}2 i[math] Substituting (to avoid all those ugly numeric constant): [math]c=-\frac32$, $d= \frac{\sqrt7}2$

$a = c +di$, $b= c -di$

Then

$a^3 +b^3$

$= c^3 +3c^2di -3cd^2 -d^3i$

$+c^3 -3c^2di -3cd^2 +d^3i$

$= 2c^3 -6cd^2$

Substuting for the constants back,

$a^3 +b^3 = \frac{-54 +6 \cdot 3 \cdot 7}8 = 9$

I'm going to guess that you made a mistake somewhere. I solved it in my previous post without any i values, and I used the value $ab=4$ and the value $a+b=(-3)$. Though for the life of me I can't find where you might have made a mistake. Perhaps I did?

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I've been but. but better late than never.

This seems a likely corollary from the binomial theorem, but I'm not sure if it would have to be proven separately that (I'm doubting a nautral number counterexample, but still):

a^n + b^n = (a+b)^n - (n(a^n-1*b)...+...(n(a*b^n-1))

so if (a+b) = s and ab = p...

(s)^n - (p(a^n-2...+...b^n-2)

However, this is where it becomes tricky as you cannot simply factor out an n, and as I was browsing through pascal's triangle, I noticed that some rows of the triangle would not factor cleanly into nice, natural numbers, of course this excludes one, since that's the coefficient of what we're interested in, so in row six, the coefficients would be 6, 15, 20, 15, 6, of which there is no natural factor (though I suppose if you really wanted a numerical answer, it would be no problem), so you can't really say you could "factor out an n", as it is so conveniently done from the case where n = 3 (if you recall your binomial theorem, where the coefficients are given by combinations, given by n!/(r!(n-r)!, the best that can be done is to factor out an "ab", since that would be greatest common factor between all the terms. Hopefully someone has a more creative insight here...

It can certainly be approached with the Newton binomial expression but, as I said above, for a given n that isn't small it needs to be done recursively. I'm not too optimistic about working out a non-recursive expression for a general n. In terms of sum and product (considering the symmetry for a-b exchange):

$a^n + b^n = s^n - \sum_{0<i\leq\frac{n}{2}}\frac{n!}{i!(n-i)!}(a^{n-2i} + b^{n-2i})p^i$

wherein each of the $\norm(a^{n-2i} + b^{n-2i})$ can of course be replaced recursively down to $\norm(a^0 + b^0) = 2$ and $\norm(a^1 + b^1) = s$.

The only remark I so far have is that, for even n the n-2i are all even and they are all odd for odd n; this determines which of the above two terminal cases the recursion will break down to.

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Beginning with the general form of the given equations

$a+b+c=0$, $ab=d$

we solve for $a$ and $b$ using the quadratic formula

$b=\frac{d}a$

$a+\frac{d}a+c=0$

$a^2+ca+d=0$

$r=4d-c^2$, $i=\sqrt{-1}$

$a=\frac{-c+\sqrt{r}i}2$, $b=\frac{-c-\sqrt{r}i}2$

when $r \gt 0$.

Using the binomial theorem, we have

$a^n = \frac{ \sum_{j=0}^n \frac{n!}{j!(n-j)!} (-c)^{n-j}(\sqrt{r}i)^j }{2^n}$

$b^n = \frac{ \sum_{j=0}^n \frac{n!}{j!(n-j)!} (-c)^{n-j}(-\sqrt{r}i)^j }{2^n}$

Note that every other term of these 2 series are additive inverses (cancel), so

$a^n+b^n= \frac{ \sum_{k=0}^{\frac{n}2} \frac{n!}{(2k)!(n-2k)!} (-c)^{n-2k}(-r)^k }{2^{n-1}}$

So, for $a+b+3=0$, $ab=4$,

$a^1 +b^1 =-3$

$a^2 +b^2 =1$

$a^3 +b^3 =9$

$a^4 +b^4 =-31$

$a^5 +b^5 =57$

etc.

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• 4 months later...

Craig's solution is the one I came up with before I realised there were more posts in the thread. :huh:

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