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Topic: **How to write a mathematical equation****Question:**
I have this assignment to prove that [2*x^3 1] will never equal a perfect cube. It seems obvious, but I can't quite see how to do it. We can't use graphs, unfortunately.
Any advice?
NOTE: Answers is removing the addition sign from the problem, sorry - forgot about that.
f(x) = 2*x^3 [plus] 1
I am trying to prove that f(x) != z^3 such that z is a real integer.
Additional Notes:
X is a non-zero positive integer.

July 17, 2019 / By Bessy

Obvious answer: 2 is not a perfect cube, so 2x^3 cannot be written as (ax)^3 where a is a whole ineteger. a^3=2, a=cbrt2, cbrt2 is not an integer. Thus 2x^3+1 is never a perfect cube. Mathematical Answer: Perfect Cube-- (a+b)^3=a^3+3a^2b+3ab^2+b^3 2x^3+1=? there is no a^2 (ie x^2) or a (ie x) term, this is only possible if b=0, which it doesnt because b^3=1, thus b=1. The (almost) perfect cube would be: (xcbrt2)^3+x*3cbrt4+x^2*3cbrt2+1= (xcbrt2+1)^3 however this is clearly not even close to that, as it is missing two whole terms. peace

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Did you like the answer? We found more questions related to the topic: **How to write a mathematical equation**

What do you mean by 2Exponent(n)? Do you mean "2 to the n", or do you mean "2 times e to the n"? I'll assume the first, and use ^ instead of "exponent". If you mean the second, the proof is similar in principle. Basis: n=4 --> Just plug in n=4 to verify the statement: 2*4+3 = 11, which is less than 2^4 = 16 Induction step: Assume you have the statement for some n, i.e. you have 2n+3 <= 2^n Now you need to prove the statement for n+1, that is, you need to prove 2(n+1)+3 <= 2^(n+1) 2(n+1)+3 = 2n + 2 + 3 = (2n + 3) + 2 <= 2^n + 2 <= 2^n + 2^n = 2^(n+1) (I used the induction hypothesis, 2n+3 <= 2^n, at the first <= sign, and the fact that 2<2n for any n>=2 at the second <= sign)

i might try this by potential of induction. (this is suggestion from somebody who has taken creation to Mathematical Exposition). i do no longer understand how your professor needs you to place in writing out this evidence. the nicely-known is at the start a base case and from there do an induction step. Assuming that that's authentic for all n in N you are able to initiate with n = a million which provides 2 < 4. here is the place my concept is provided in. We already understand from the backside case that 2 < 4. Then 4n + 2 < 4n + 4 for all n in N. Iff (4n + 2)/(n + a million) < 4 enable P(n+a million):= "(2n+2, n+a million) <= 4^(n+a million)" Then 2(2n+a million)/(n+a million) *(2n,n) = (4n + 2)/(n + a million) *(2n,n) < 4*4^(n) = 4^(n+a million) it incredibly is the assumption, with any luck you are able to end the write up. I used P(n) the inductive hypothesis. Goodluck!

I might comment that the statement is false. f(0) = 2*0^3 + 1 = 1 = 1^3, which is a perfect cube. So did you want x to be a positive integer?

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I'd like to help but want to know whats happening with the 1 at the end? If I'm right then you want to prove that for y = 2x^3 + 1 then y cannot equal z^3 (z is in the reals) Or is this different notation I am not familiar with?

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It's like saying "Well, there's books and movies and stuff written about Spiderman, and New York City exists, and crime exists, and spiders exist, so that must mean Spiderman exists!". Christian apologist logic for you.

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