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Suppose that we have a solution of distinct reals (a,b,c). Let S=a2+b2+c2.
Then, we have a3+3a2+25=3(a2+b2+c2)=3S, and similar equations for b and c.
Since these are distinct reals, we know that the roots to the cubic equation X3+3X2+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=(a2+b2+c2)=(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 x3+3x2−2=0 to obtain that (a,b,c)=(−1,−1−3,−1+3).
first subtract the equation simultaneously to get
⎩⎪⎨⎪⎧a3−b3=−3(a2−b2)b3−c3=−3(b2−c2)c3−a3=−3(c2−a2)
since all variable are distinct we can divide both side by (a-b),(b-c),(c-a) respectively. we get
⎩⎪⎨⎪⎧a2+ab+b2+3(a+b)=0b2+bc+c2+3(b+c)=0c2+ca+c2+3(c+a)=0
subtract any two equation to get
a+b+c=−3
add all the equations to get
2(a2+b2+c2)+ab+bc+ca+6(a+b+c)=0
using the facta2+b2+c2=(a+b+c)2−2(ab+bc+ca)=9−2(ab+bc+ca)ab+bc+ca=0
adding all the original equations
a3+b3+c3=6(a2+b2+c2)−75=−21
using the identity
a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc=−27+3abc
so we have
−27+3abc=−21→abc=2
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Top NewestSuppose that we have a solution of distinct reals (a,b,c). Let S=a2+b2+c2.
Then, we have a3+3a2+25=3(a2+b2+c2)=3S, and similar equations for b and c.
Since these are distinct reals, we know that the roots to the cubic equation X3+3X2+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=(a2+b2+c2)=(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 x3+3x2−2=0 to obtain that (a,b,c)=(−1,−1−3,−1+3).
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good simple approach!
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Hint: What can we say about a3+3a2+25?
The next hint is in white text, so you will need to Toggle Latex.
Hint: For a given solution set, consider f(x)=x3+3x2+25−3(a2+b2+c2).What are the roots?
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@Calvin Lin Please help me out with the problem!!!!
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Which problem? This one? If so, see the hint.
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But otherwise, (since the above is already a spoiler space), the second hint says
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first subtract the equation simultaneously to get ⎩⎪⎨⎪⎧a3−b3=−3(a2−b2)b3−c3=−3(b2−c2)c3−a3=−3(c2−a2) since all variable are distinct we can divide both side by (a-b),(b-c),(c-a) respectively. we get ⎩⎪⎨⎪⎧a2+ab+b2+3(a+b)=0b2+bc+c2+3(b+c)=0c2+ca+c2+3(c+a)=0 subtract any two equation to get a+b+c=−3 add all the equations to get 2(a2+b2+c2)+ab+bc+ca+6(a+b+c)=0 using the facta2+b2+c2=(a+b+c)2−2(ab+bc+ca)=9−2(ab+bc+ca) ab+bc+ca=0 adding all the original equations a3+b3+c3=6(a2+b2+c2)−75=−21 using the identity a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc=−27+3abc so we have −27+3abc=−21→abc=2
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Awesome solution...(+1) one of the best.
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The answer is abc=2
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can you post your solution?
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This is from NMTC 2015 lvl 2
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you are absolutely correct.
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What is the sum of 1!+2!+3!+4!+.....+n!
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why do you want that?
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Actually, there is no neat closed form expression for ∑n=1kn!.
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 ∑n=1kn⋅n! which can be described by simple expressions.
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