## If M = square root 4 + cube root 4 + fourth root of 4, then

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RonPurewal
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### Re: If M = square root 4 + cube root 4 + fourth root of 4, then

nope. that's ... not a thing at all, actually. there's no way to simplify a sum of powers.

there's nothing wrong with having that idea in your head--from time to time, we all toss around ideas that turn out to be wrong--but it's very easy to disprove that idea with simple examples.
e.g., 1^2 + 1^3 + 1^4 is actually 3, but your method of figuring would give 1^9 = 1.
RonPurewal
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### Re: If M = square root 4 + cube root 4 + fourth root of 4, then

more generally--if there were any way to simplify a sum of powers, then polynomials wouldn't exist!

you know, those things like x^2 + x + 6 that you spent so much time factoring in high school.
if you could simplify something like x^2 + x^1 into a single term, then those things wouldn't be things.
NicholasW85
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### Re: If M = square root 4 + cube root 4 + fourth root of 4, then

I understand the estimation theory, but why couldnt you have removed a common base of 2 from all the terms in the problem. See below for how I thought of the problem:

sqrt4 = 2
cbrt4 = 2^(1/3)
fthrt4 = 2^(1/4)

Since they all have a 2 as the base then I figured I could re-write the equation as:

2*(1 + 1^1/3 + 1^1/4)

Calculating the numbers inside the parentheses is easy since they're all equal to 1, right? So you get:

2*(1+1+1) = 6 (greater than 4)

I may be missing something so please do let me know!

Thanks!
RonPurewal
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### Re: If M = square root 4 + cube root 4 + fourth root of 4, then

that's definitely incorrect.
√4 is equal to 2, so, the cube root of 4 and the fourth root of 4 are clearly smaller than 2. therefore, the whole sum is most certainly less than 6.

the problem is that you're trying to factor out integers that you can't factor out.

for instance, √10 equals √2 times √5... but, what you're doing would be like trying to make √10 equal 2 times √5, or 5 times √2. nope. can't do that.