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I just realised that if we expand any of the non-linear expression with power of 3 or more we can't stop expanding them until we are dead.

So for example:

Expansion:
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$

Grouping:
$(a+b+c)^2 = \textbf{(a + b + c)}^{\textbf 2} + 2(ab+ bc + ac)$

The bold part in the very above equation has an expression that can be again expanded to $a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$ and so that will go on forever.

Can anyone tell me if I am wrong anywhere or is this true?

Tim Ratigan
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Mohammad Areeb Siddiqui
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1 Answers1

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There is an important principle in algebra, that the order of operations must be observed.

$$(a+b+c)^2$$

requires you to perform the additions first, followed by the squaring operation.

$$a^2+b^2+c^2$$

requires you to perform the squarings first, and then the additions. You cannot switch between them, generally. To demonstrate why this is true, notice that

$$ 1^2 + 2^2 + 3^2 = 14 \neq 36 = (1+2+3)^2 $$ And so, even with this simple example, you can see that $a^2+b^2+c^2\neq(a+b+c)^2$.

Grouping involves recognition of where an operation distributes over another operation. $ab+ac = a(b+c)$. This works because multiplication distributes over addition. However, the squaring operation does not distribute over addition - instead, it distributes over multiplication. So $(ab)^2 = a^2b^2$. But $(a+b)^2=a^2+2ab+b^2\neq a^2+b^2$ (unless $2ab=0$, which only happens when $a=0$ or $b=0$).

Your expression for $(a+b+c)^3$ is also in error, the right hand side is the expansion of $(a+b+c)^2$. But I suspect that is a typo.

Glen O
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