Why can't you multiply the exponents when you have addition involved?

Question

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 :)

Answer 1

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.

Answer 2

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.

Answer 3

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 $a^2+b^2$.

Answer 4

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 $2ab$.