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For example, $(8^{\frac{1}{3}} + 27^{\frac{1}{3}})^{2}$ why can't you make this $8^{\frac{2}{3}} + 27^{\frac{2}{3}}$?

Please explain in a very simple way, thank you :)

Matthew Cassell
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Little Black Dog
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    You could try it with simpler numbers. What are the values of $(1^{1/3}+1^{1/3})^2$ and $1^{2/3}+1^{2/3}$ respectively? Another way to answer it is by asking instead, why should it be true? – Harald Hanche-Olsen Dec 01 '14 at 07:21
  • Thank you, you're right, I can see that it doesn't work, but what about one of the laws of exponents which says that you have to multiply the exponents inside the brackets with the exponent outside the bracket? or is that only when the terms are being multiplied? Sorry if this is a stupid question – Little Black Dog Dec 01 '14 at 07:31
  • Indeed, only when they are multiplied. The basic reason it works then is that exponentiation arises from repeated multiplication – it is exactly this, when the exponents are natural numbers. When you exponentiate a sum, things are different because addition and multiplication don't mix all that well. – Harald Hanche-Olsen Dec 01 '14 at 07:45

4 Answers4

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Just a numerical example of why $(a+b)^2 \neq a^2 + b^2$: $$ 25 = (5)^2 = (2+3)^2 \neq 2^2 + 3^3 = 4+9 = 13 $$ This also applies if there are exponents on $a$ and $b$, like $\frac{1}{3}$ in your case.

Jean-Claude Arbaut
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Mike Pierce
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As to why $(a+b)^2\neq a^2+b^2$, consider $(a+b)^2$ to be the area of a square of side $a+b$. And likewise, $a^2$ and $b^2$ are the areas of two smaller squares of side $a$ and $b$.

Now, you can inscribe these smaller squares in the larger one, at opposite corners. And there is much room left: two rectangles of sides $a,b$, that is, total area $2\times a\times b$ is missing.

Thus, you have $(a+b)^2=a^2+b^2+2ab$. When both sides $a,b$ are positive, this is not equal to $a^2+b^2$.

Jean-Claude Arbaut
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Consider a similar question of the same type. I have $1 + (2 \times 3)$. Why isn't this equal to $(1 + 2) \times (1 + 3)$?

In one case, you did a bunch of operations on some numbers. In the other case, you did some different operations on some numbers. If the answer was always the same regardless of which operations you did, why would we bother with having different operations at all?

Sometimes different operations do give you the same answer. For example, $a\times (b+c)$ always gives the same answer as $(a\times b) + (a\times c)$. This is remarkable when it happens, and there is always a deep reason. You should not be surprised that it does not happen every time.

MJD
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  • Thanks, I know this should be obvious... but I don't get it. I thought that the laws of exponents says to multiply the exponents inside the brackets with the exponent outside the brackets.. Does that only hold if the terms are being multiplied? For some reason, this is not obvious to me :( – Little Black Dog Dec 01 '14 at 07:33
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What you're doing is squaring a binomial. When you square a binomial like $(a+b)^2$, you get $a^2+2ab+b^2$. So you do get $a^2 + b^2$, but you also get another term, $2ab$.

To understand this, consider the FOIL method. Write $(a+b)^2$ as $(a+b)(a+b)$. You'll end up multiplying the $a$ in the first binomial with the $b$ in the second binomial, and also multiplying the $b$ in the first binomial with the $a$ in the second binomial. So two terms will be $ab$, and they'll add together to make $2ab$.

Kyle Delaney
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