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$a + b +c = 17$

$a^2 + b^2 + c^2 = 101$

$a^3 + b^3 + c^3 = 623$

How does one go about solving this?

Thanks

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  • $\begingroup$ Welcome to MSE! For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – Cortizol
    Dec 2, 2013 at 13:05
  • $\begingroup$ tried $(a+b+c)^2=??$ and $(a+b+c)^3=???$ $\endgroup$
    – user87543
    Dec 2, 2013 at 13:05
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/27394/…. $\endgroup$
    – lhf
    Dec 2, 2013 at 13:07
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    $\begingroup$ @Cortizol - Please don't bombard first time users with a barrage of links about formatting, particularly when the question is unambiguous as is. That's hardly a warm welcome to a first-time user. $\endgroup$
    – amWhy
    Dec 2, 2013 at 13:09

2 Answers 2

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Hint:

Expand $(a + b + c)^2$ and $(a + b + c)^3$, and use the three equations you are given.

$$(a+b+c)^2 = (17)^2 = \underbrace{\color{blue}{a^2 + b^2 + c^2}}_{101} +2(\bf{ab + ac + bc})$$

$$\begin{align} (a + b+ c)^3 = (17)^3 & = \underbrace{\color{red}{a^3 + b^3 + c^3}}_{623} + 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2) \\ \\&= 623 + 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 + \color{green}{abc - abc})\\ \\ & = 623 + 3\Big(\underbrace{(a + b + c)}_{17}{\bf(ab + ac + bc)}- \color{green}{\bf abc}\Big)\end{align}$$

Spoiler:

Try the triple of numbers $4, 6, 7$.

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  • $\begingroup$ Love the spoiler :) $\endgroup$
    – Mikasa
    Dec 2, 2013 at 13:34
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my weaknesses show up particularly when doing tedious calculations, so as a penance, just this once, i will bludgeon my way through this, probably generating more heat than light in the process!

we have: $$a+b=17-c$$ thus $$(a+b)^2 = 289 -34c+c^2$$ and $$(a+b)^3 = 4913 - 867c+51c^2 -c^3$$

but $$(a+b)^2 = a^2+b^2 + 2ab = 101 - c^2 +2ab$$ giving $$ab = 94 -17c +c^2$$

likewise $$(a+b)^3 = a^3+b^3 +3ab(a+b) \\ = 623 -c^3 +3(94-17c+c^2)(17-c) \\ =5417 -1149c +102c^2 -4c^3$$ so we have $$ 5417 -1149c +102c^2 -4c^3 = = 4913 -867c+51c^2 -c^3$$ i.e. $$504 -282c+51c^2 -3c^3 =0$$ or $$c^3-17c^2+94c-168 =0$$ or $$(c-4)(c-6)(c-7) = 0$$

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