Question

a3 + b3 = C3 tell me what numbers can we include to it..?

Answer 1

its actually \[a ^{3} + b ^{3} = c ^{3}\]

Answer 2

This was a futurama joke. Google found this http://www.mathpuzzle.com/twocubes.txt

Answer 3

zero works ....

Answer 4

No non-zero integers...

Answer 5

This is actually a trick question. It's a variation of Fermat's Last Theorem. It states that for x<2, a^x+b^x=c^x does not exist. Fermat didn't write down his proof, so no one knew how he figured it out. It went unsolved for over 300 years. Not too long ago, a man (I forget his name) finally figured it out. We can now prove that there is not a single set of numbers that you can plug into that equation with all three exponents being the same (with an exponent over two).

Answer 6

Andrew Wiles

Answer 7

Side note time! Thanks for the medals guys! I wasn't expecting such a response! I love sharing my random extra math knowledge with people, and this is the first place I've actually gotten a good response to it. Also, if you're interested in this topic, I highly suggest Simon Singh's book, Fermat's Enigma. I'm about halfway through it, and I love it. Then again, I absolutely love Simon Singh in general. Anyway, it's an extremely interesting topic and I really think it's worth learning more about. Thanks for listening to me ramble in my nerdy way for a bit!

Answer 8

There are infinite solutions.

Answer 9

No Parth, there are no solutions. This is a Fermat's last theorem question.

Answer 10

@singlesixx do you see the asker mentioning anywhere that they wanted integral solutions?

Answer 11

\[0^3 + 0^3 = 0^3\]Booyah.

Answer 12

no we can't

Answer 13

\[1^3 + 0^3 = 1^3\]BOOYAH

Answer 14

no dude

Answer 15

@singlesixx The original poster hasn't explicitly asked us for integral non-zero distinct solutions. That's three things you're assuming :-)

Answer 16

it can't be small numbers because global tried this....

Answer 17

@singlesixx Also, we're talking about \(x > 2\). You wrote \(x<2\)

Answer 18

@UnkleRhaukus But we gotta love Andrew Wiles, eh? :-)

Answer 19

\[(i^{2/3})^3 + (5^{1/3})^3 = (2^{2/3})^3\]

Answer 20

lol

Answer 21

Parth, honey, cut it out. You're just mad because I know what I'm talking about. I did not mention integral solutions. I did make a mistake in saying x<2 instead of x>2, but my point was still clear. 0, 1, and 2 are the exponents that are generally accepted to work. Yes, you can get around that by using 1 or 0 as the base, but nobody really thinks of that as a solution. I know what I'm talking about, alright? This is one of my favorite theorems and I know a lot about it. You're trying to undermine me to make yourself look better and it's childish.

Answer 22

how old exactly do you think the great PK is? @singlesixx

Answer 23

I have no way of knowing how old he is. Frankly, it doesn't matter.

Answer 24

13

Answer 25

Okay, why does that matter? I know several 13 year olds that are pretty mature. I hope you aren't using your age as an excuse. I'm only three years older than you.

Answer 26

mmmmm...

Answer 27

\(a^n+b^n=c^n\) as no integer solutions if \(n\geq 3\)
it is fermat's last theorem, but the case for \(n=3\) is not too hard
case \(n=4\) is even easier
general proof would require probably 10 years of study to being to understand the structure of the proof i imagine

Answer 28

\[\large 0^n + 0^n = 0^n\]

Answer 29

your question is not wrong

Answer 30

for example i will give you one solution x=1/{1/2}^(1/3), y=1/{1/2}^1/3 and z=1/{1/4}^1/3 these are real values there are also so many real vaues that can satisfy your equation.

Answer 31

i mean x=a, y=b and z=c

Answer 32

i asked numbers dude....

Answer 33

just use calculator

Answer 34

I would say that any positive integer can be included in this equation. This is simply say that the sum of two cubes is equal to another cube. Any positive number can have a cube root, so?

Answer 35

not to bump this but it showed up under 'open questions' and I felt compelled to respond... @singlesixx we're all glad that you have 'extra' math knowledge but you needn't take being proven wrong so personally. Although initially I was going to reward you a medal, reading your immature responses to PK's valid challenges (mostly because of the ambiguity with which you stated Fermat's last theorem) changed my mind. Ironically enough you called his behavior 'childish'... you should avoid coming across as arrogant when this website has several mathematicians far better than both you and I :-p

Answer 36

still i didn't got it.. :(

Answer 37

Read this
http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

Answer 38

great..

Answer 39

why is this bumped?