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{a3=3(b2+c2)25b3=3(c2+a2)25c3=3(a2+b2)25\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,ba,b and cc be distinct real numbers satisfying the system of equations above. Find the product abcabc.

Note by Ayush G Rai
3 years, 5 months ago

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Suppose that we have a solution of distinct reals (a,b,c) (a,b,c) . Let S=a2+b2+c2 S = a^2 + b^2 + c^2 .
Then, we have a3+3a2+25=3(a2+b2+c2)=3S a^3 + 3a^2 + 25 = 3 (a^2 + b^2 + c^2 ) = 3S , and similar equations for bb and cc.

Since these are distinct reals, we know that the roots to the cubic equation X3+3X2+253S=0 X^3 + 3X^2 + 25 - 3S = 0 must be a,b,ca, b, c .
Hence, by vieta's formula, this tells us a+b+c=3,ab+bc+ca=0 a+b+c = -3, ab+bc+ca = 0 .

Thus, S=(a2+b2+c2)=(a+b+c)22(ab+bc+ca)=9 S = (a^2+b^2+c^2) = (a+b+c)^2 - 2(ab+bc+ca) = 9 . Finally, using the constant coefficient, abc=(253S)=(2527)=2 abc = - ( 25 - 3S ) = - (25 - 27) = 2 .


Note: We can now solve x3+3x22=0 x^3 + 3x^2 -2 = 0 to obtain that (a,b,c)=(1,13,1+3) (a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} ) .

Calvin Lin Staff - 3 years, 5 months ago

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good simple approach!

Ayush G Rai - 3 years, 5 months ago

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Hint: What can we say about a3+3a2+25 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 f(x)=x3+3x2+253(a2+b2+c2).What are the roots?\color{#FFFFFF} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}}

Calvin Lin Staff - 3 years, 5 months ago

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@Calvin Lin Please help me out with the problem!!!!

Ankit Kumar Jain - 3 years, 5 months ago

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Which problem? This one? If so, see the hint.

Calvin Lin Staff - 3 years, 5 months ago

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@Calvin Lin How to toggle latex??

Ankit Kumar Jain - 3 years, 5 months ago

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@Ankit Kumar Jain On browser, click on your profile icon

But otherwise, (since the above is already a spoiler space), the second hint says

For a given solution set, consider f(x)=x3+3x2+253(a2+b2+c2). f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). What are the roots?

Calvin Lin Staff - 3 years, 5 months ago

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@Calvin Lin Sir it hence means that ABC = 2

Ankit Kumar Jain - 3 years, 5 months ago

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first subtract the equation simultaneously to get {a3b3=3(a2b2)b3c3=3(b2c2)c3a3=3(c2a2)\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 {a2+ab+b2+3(a+b)=0b2+bc+c2+3(b+c)=0c2+ca+c2+3(c+a)=0\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=3a+b+c=-3 add all the equations to get 2(a2+b2+c2)+ab+bc+ca+6(a+b+c)=02(a^2+b^2+c^2)+ab+bc+ca+6(a+b+c)=0 using the facta2+b2+c2=(a+b+c)22(ab+bc+ca)=92(ab+bc+ca)a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca) ab+bc+ca=0ab+bc+ca=0 adding all the original equations a3+b3+c3=6(a2+b2+c2)75=21a^3+b^3+c^3=6(a^2+b^2+c^2)-75=-21 using the identity a3+b3+c3=(a+b+c)(a2+b2+c2abbcca)+3abc=27+3abca^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=21abc=2-27+3abc=-21\to abc=2

Aareyan Manzoor - 3 years, 5 months ago

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Awesome solution...(+1) one of the best.

Ayush G Rai - 3 years, 5 months ago

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The answer is abc=2abc=2

Janardhanan Sivaramakrishnan - 3 years, 5 months ago

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can you post your solution?

Ayush G Rai - 3 years, 5 months ago

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This is from NMTC 2015 lvl 2

Ankit Kumar Jain - 3 years, 5 months ago

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you are absolutely correct.

Ayush G Rai - 3 years, 5 months ago

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What is the sum of 1!+2!+3!+4!+.....+n!

Bunneng Nath - 3 years, 5 months ago

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why do you want that?

Ayush G Rai - 3 years, 5 months ago

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Actually, there is no neat closed form expression for n=1kn!\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 n=1knn!\sum_{n=1}^{k}\,n \cdot n! which can be described by simple expressions.

Aditya Sky - 3 years, 5 months ago

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