Question

\(a^3 + b^3 + c^3 = 0 \) then find : \((a+b+c)_{(\textbf{min.})}\)

Answer 1

\(a^3 + b^3 + c^3 = 0 \) then find : \((a+b+c)_{(\textbf{min.})}\)

Justify your each step also.

Answer 2

@uri
sorry I edited the question, so you have to refresh your page.

Answer 3

Remember I will ask for the proof of each and every step.

Answer 4

\(\textbf{a,b and c} \in \mathbb{R}\)

Answer 5

\[a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca) + 3abc\]So \(abc = 0\) and \(a + b + c = 0\).

Answer 6

Hmm... no

Answer 7

good start point parth :)

Answer 8

@mukushla Is that right?

Answer 9

nope :)

Answer 10

I think it should be \(abc = 0\) OR \(a + b + c = 0\).

Answer 11

but the identity is good

Answer 12

:)

Answer 13

OR \(a^2 + b^2 + c^2 -ab-bc-ca =0\).

Answer 14

I wonder if that even helps.

Answer 15

\[a^3 + b^3 + c^3 = (a + b)(a^2 - ab + b^2) + c^3\]How about this one, maybe?

Answer 16

Nope, why can you say abc=0 or a+b+c=0? It's not true from that identity.

Answer 17

It's actually something like this:

either

\(abc = 0\) and \(a + b + c = 0\)

or

\(abc = 0\) and \(a^2+b^2+c^2−ab−bc−ca=0\).

or

\(abc = 0\) and \(a + b + c = 0\) and \(a^2+b^2+c^2−ab−bc−ca=0\).

Answer 18

abc can be positive and (a+b+c) can be negative or vice versa, a^3+b^3+c^3 can still be zero.

Answer 19

hmm

Answer 20

You're right.

Answer 21

;)

Answer 22

Example?

Answer 23

Have you learnt Lagrange multipliers @mathslover ?

Answer 24

:-|||

Answer 25

@drawar Are you Indian?

Answer 26

Nope, why do you come up with that question?

Answer 27

No but you can put your opinion here.

Answer 28

Nothing, your name seemed Indian.

Answer 29

Haha, just a random name, it has nothing to do with my real name.

Answer 30

it's a method of finding the maximum/minimum values of functions subject to a number of constraints

Answer 31

oh, yeah I remember I saw its name somewhere and turned over the page :P

Answer 32

Hmmm

Answer 33

Any easier method you have?

Answer 34

(a+b+c)/3 >= (abc)^1/3

and
(a^3 + b^3 + c^3)/3 >= abc

Answer 35

is 0 the ans ?

Answer 36

Yes ... 0 is the answer shubham .

Answer 37

well is my my method satisfactory ?

Answer 38

It is easy that a+b+c will have 0 as the min. value but how will u prove that?

Answer 39

I used AM>=GM

Answer 40

The expressions are right but how can u say that it is 0?

Answer 41

(a+b+c)/3 >= (abc)^1/3

and
(a^3 + b^3 + c^3)/3 >= abc

In second eqn,
0 >= abc

so max value of abc is 0
and a+b+c>= 3 (abc)^1/3

when abc is max, a+b+c is min
and hence the answer.

Answer 42

\(\color{blue}{\textbf{EXCELLENT!}} \quad \ddot \smile\)

Answer 43

Great work. @shubhamsrg

Answer 44

Wel, glad I could contribute

Answer 45

well*

Answer 46

:)

Answer 47

You deserve medals bro. Good work.