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# Can someone prove that for the equation f(x) = 2*x^3 1 that f(x) can't equal a perfect cube?

Topic: How to write a mathematical equation
July 17, 2019 / By Bessy
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.

## Best Answers: Can someone prove that for the equation f(x) = 2*x^3 1 that f(x) can't equal a perfect cube?

Afton | 7 days ago
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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We found more questions related to the topic: How to write a mathematical equation

Originally Answered: Prove that 2n+3 equal or less than 2Exponent(n) for all n equal or greater than 4. thanks :)?
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)
Originally Answered: Prove that 2n+3 equal or less than 2Exponent(n) for all n equal or greater than 4. thanks :)?
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!

Tiarnan
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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