200 Giveaway

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

I'm sorry to say but your answer is wrong.

---

15-11-2015, 05:57 PM

NSA Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

You put the numbers wrongly in the calculator. It must be like this:(a)^3+(b)^3+©^3=33 not a^3+b^3+^3=33.

15-11-2015, 06:05 PM

Gnar Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

I'm sorry to say but your answer is wrong.

---

15-11-2015, 05:57 PM

NSA Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

You put the numbers wrongly in the calculator. It must be like this:(a)^3+(b)^3+©^3=33 not a^3+b^3+^3=33.

What calculator do you use, it looks cool

Lol this question can't be solved

It is still not known whether 33,42 and 74 can be written as the sum of three cubes.

In this case there is no solution and this giveaway will become fake.

Aoki

15-11-2015, 06:24 PM

Darksider Wrote: Lol this question can't be solved

It is still not known whether 33,42 and 74 can be written as the sum of three cubes.

In this case there is no solution and this giveaway will become fake.

Aoki

He insisted it can be solved though.

15-11-2015, 06:22 PM

NSA Wrote:

15-11-2015, 06:05 PM

Gnar Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

I'm sorry to say but your answer is wrong.

---

15-11-2015, 05:57 PM

NSA Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

You put the numbers wrongly in the calculator. It must be like this:(a)^3+(b)^3+©^3=33 not a^3+b^3+^3=33.

What calculator do you use, it looks cool

Here is the calculatorhttp://[disallowed url shortening site]/IqT6zt

15-11-2015, 06:32 PM

Gnar Wrote:

15-11-2015, 06:22 PM

NSA Wrote:

15-11-2015, 06:05 PM

Gnar Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

I'm sorry to say but your answer is wrong.

---

15-11-2015, 05:57 PM

NSA Wrote:

15-11-2015, 09:10 AM

education Wrote: a3 + b3 + c3 = 33


Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.

1 + 8 + 27 = 36


This sum is close to 36.

-2 + 8 + 27 = 33

3(-2)3 + 23 + 33 = 33


Therefore,

a = 3(-2) = -1.2299

b = 2

c = 3

You put the numbers wrongly in the calculator. It must be like this:(a)^3+(b)^3+©^3=33 not a^3+b^3+^3=33.

What calculator do you use, it looks cool

Here is the calculatorhttp://[disallowed url shortening site]/IqT6zt

Just kidding,
here it is:http://web2.0calc.com/

Jokes on you, I like that song. :yus:

15-11-2015, 06:24 PM

Darksider Wrote: Lol this question can't be solved

It is still not known whether 33,42 and 74 can be written as the sum of three cubes.

In this case there is no solution and this giveaway will become fake.

Aoki

It is none of my problem ifit isknown or not.
How is it fake? If you write answer that calculator shows is right then you get the money?

NSA lol this is funny

http://prntscr.com/932637

15-11-2015, 06:39 PM

Believer Wrote: NSA lol this is funny

http://prntscr.com/932637

not me. I just wanted to use the same calculator as him.