\[\large{ \begin{cases} a^3 = 3(b^2+c^2)-25 \\ b^3=3(c^2+a^2)-25 \\ c^3=3(a^2+b^2)-25 \end{cases}} \]
Let \(a,b\) and \(c\) be distinct real numbers satisfying the system of equations above. Find the product \(abc\).
\[\large{ \begin{cases} a^3 = 3(b^2+c^2)-25 \\ b^3=3(c^2+a^2)-25 \\ c^3=3(a^2+b^2)-25 \end{cases}} \]
Let \(a,b\) and \(c\) be distinct real numbers satisfying the system of equations above. Find the product \(abc\).
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Ayush Rai
1 year, 3 months ago
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Top NewestSuppose that we have a solution of distinct reals \( (a,b,c) \). Let \( S = a^2 + b^2 + c^2 \).
Then, we have \( a^3 + 3a^2 + 25 = 3 (a^2 + b^2 + c^2 ) = 3S \), and similar equations for \(b\) and \(c\).
Since these are distinct reals, we know that the roots to the cubic equation \( X^3 + 3X^2 + 25 - 3S = 0 \) must be \(a, b, c \).
Hence, by vieta's formula, this tells us \( a+b+c = -3, ab+bc+ca = 0 \).
Thus, \( S = (a^2+b^2+c^2) = (a+b+c)^2 - 2(ab+bc+ca) = 9 \). Finally, using the constant coefficient, \( abc = - ( 25 - 3S ) = - (25 - 27) = 2 \).
Note: We can now solve \( x^3 + 3x^2 -2 = 0 \) to obtain that \( (a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} ) \). – Calvin Lin Staff · 1 year, 2 months ago
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good simple approach! – Ayush Rai · 1 year, 2 months ago
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Hint: What can we say about \( a^3 + 3a^2 + 25 \)?
The next hint is in white text, so you will need to Toggle Latex.
Hint: For a given solution set, consider \(\color{white} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}} \) – Calvin Lin Staff · 1 year, 3 months ago
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@Calvin Lin Please help me out with the problem!!!! – Ankit Kumar Jain · 1 year, 3 months ago
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Which problem? This one? If so, see the hint. – Calvin Lin Staff · 1 year, 3 months ago
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How to toggle latex?? – Ankit Kumar Jain · 1 year, 3 months ago
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On browser, click on your profile icon
But otherwise, (since the above is already a spoiler space), the second hint says
– Calvin Lin Staff · 1 year, 3 months agoLog in to reply
Sir it hence means that ABC = 2 – Ankit Kumar Jain · 1 year, 3 months ago
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first subtract the equation simultaneously to get \[\begin{cases} a^3-b^3=-3(a^2-b^2)\\ b^3-c^3=-3(b^2-c^2)\\ c^3-a^3=-3(c^2-a^2) \end{cases}\] since all variable are distinct we can divide both side by (a-b),(b-c),(c-a) respectively. we get \[\begin{cases} a^2+ab+b^2+3(a+b)=0\\ b^2+bc+c^2+3(b+c)=0\\ c^2+ca+c^2+3(c+a)=0\\ \end{cases}\] subtract any two equation to get \[a+b+c=-3\] add all the equations to get \[2(a^2+b^2+c^2)+ab+bc+ca+6(a+b+c)=0\] using the fact\(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca)\) \[ab+bc+ca=0\] adding all the original equations \[a^3+b^3+c^3=6(a^2+b^2+c^2)-75=-21\] using the identity \[a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=-27+3abc\] so we have \[-27+3abc=-21\to abc=2\] – Aareyan Manzoor · 1 year, 2 months ago
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Awesome solution...(+1) one of the best. – Ayush Rai · 1 year, 2 months ago
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The answer is \(abc=2\) – Janardhanan Sivaramakrishnan · 1 year, 3 months ago
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can you post your solution? – Ayush Rai · 1 year, 3 months ago
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This is from NMTC 2015 lvl 2 – Ankit Kumar Jain · 1 year, 3 months ago
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you are absolutely correct. – Ayush Rai · 1 year, 3 months ago
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What is the sum of 1!+2!+3!+4!+.....+n! – Bunneng Nath · 1 year, 3 months ago
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Actually, there is no neat closed form expression for \(\sum_{n=1}^{k}\,n!\).
There are some formulae but you would need to know about exponential integrals and several other advance concepts for understanding them.
However, there are some series involving factorials like \(\sum_{n=1}^{k}\,n \cdot n!\) which can be described by simple expressions. – Aditya Sky · 1 year, 2 months ago
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why do you want that? – Ayush Rai · 1 year, 3 months ago
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