Question

Nice problem with a nice solution.

Answer 1

Yes the problem is?

Answer 2

Solve\[\alpha + \beta + \gamma=0\]\[\alpha \beta + \beta \gamma + \alpha \gamma = 0\]\[\alpha\beta\gamma = 0\]

Answer 3

isn't that the cubic?

Answer 4

your white too :D who was white first???

Answer 5

inky

Answer 6

btw, inky is asian.

Answer 7

I'm not white, i'm invisible. There is a difference

Answer 8

lol i wasn't tlking about skin lol

Answer 9

x^3 + ( a + b + c)*x^3 + (ab + ac + bc)*x + abc

Answer 10

hm thats why i put this because i was invisible

Answer 11

Bye guys, perl has done it.

Answer 12

sad

Answer 13

More problems please bro @ParthKohli

Answer 14

now @Here_to_Help15 , I'm invisible. You are white. Check our profile pics. THere is a difference

Answer 15

@perl Explain, explain pleaseeeeee~~~

Answer 16

(x-a)(x-b)(x-c) callisto

Answer 17

http://en.wikipedia.org/wiki/Vieta%27s_formulas @Callisto

Answer 18

at least I think @ParthKohli am i rite

Answer 19

Yup.

Answer 20

Still confused.
Btw, it would be x^3 + ( a + b + c)*x^2 + (ab + ac + bc)*x + abc then

Answer 21

That'd reduce our cubic to \(p(x) = x^3\) aaaaaaaaand.

Answer 22

hahha you caught the two calliso :3

Answer 23

The only possible value of \(\alpha, \beta , \gamma\) is 0.

Therefore \(\alpha = \beta = \gamma = 0\)

Answer 24

This is not nice.. This is trivial...

Answer 25

lol @Callisto yearn for @foolformath problems?

Answer 26

Posting one foolish problem right away.

Answer 27

the solutions to
a+b+c = m
ab +ac + bc = p
abc =q

are the same
as the roots to

x^3 + p*x^2 +q*x + c = 0

Answer 28

and since m=0 , p = 0 , q = 0
we have , the solutions are
the roots of
x^3 + 0*x^2 + 0*x + 0 = 0
x^3 = 0
x = 0

so the only solution is zero

Answer 29

typo *

Answer 30

That's clear, thanks! :)

Answer 31

the solutions to
a+b+c = m
ab +ac + bc = p
abc =q

are the same as the roots of
x^3 + m*x^2 +p*x + q = 0

Answer 32

\[x^3 - mx^2 + px - q = 0\]

Answer 33

I got your idea :)

Answer 34

oh good point, the sign alternates

Answer 35

Idea matters LOL
I was wondering why

the solutions to
a+b+c = m
ab +ac + bc = p
abc =q
are the same as the roots to
x^3 + p*x^2 +q*x + c = 0

But I understand it now. Thanks!

Answer 36

yes, i had to think about it :)

Answer 37

For recap later:

Solve:
a + b + c = 0
ab + bc + ac = 0
abc = 0



Let a, b, c be the roots of a cubic equation in unknown x. Then we have (x-a)(x-b)(x-c) = 0.
Consider (x-a)(x-b)(x-c) = 0, expand the left side, we get
\(x^3 - (a+b+c)x^2 + (ab + bc + ac)x - abc = 0\)
Substitute the values of the given equations into this polynomial equation, we have
\( x^3 = 0\)
Hence, we have x = 0 as the only solution.
Therefore, we only have the trivial solution a = b = c = 0 to the given system.