this post was submitted on 27 Nov 2025
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So is 3xy, according to that textbook. That doesn't mean 3xy^2^ is 9*y^2^*x^2^. The power only applies to the last element... like how (8)2^2^ only squares the 2.
Four separate textbooks explicitly demonstrate that that's how a(b)^c^ works. 6(ab)^3^ is 6(ab)(ab)(ab), not (6ab)(6ab)(6ab). 3(x+1)^2^ for x=-2 is 3, not 9. 15(a-b)^3^x^2^ doesn't involve coefficients of 3375. 2(x-b)^2^ has a 2b^2^ term, not 4b^2^. If any textbook anywhere shows a(b)^c^ producing (ab)^c^, or x(a-b)^c^ producing (xa-xb)^c^, then reveal it, or shut the fuck up.
2(ab)^2^ is 2(ab)(ab) the same way 6(ab)^3^ is 6(ab)(ab)(ab). For a=8, b=1, that's 2*(8*1)*(8*1).
That's right
That's right. It means 3abb=(3xaxbxb)
Factor yes, hence the special rule about Brackets and Exponents that only applies in that context
It doesn't do anything, being an invalid syntax to follow brackets immediately with a number. You can do ab, a(b), but not (a)b
Yep, as opposed to 6(a+b), which is (6a+6b)
No it isn't. See previous point. Do we have an a(b+c), yes we do. Do we have an a(bc)²? No we don't.
No, it has a a(b-c) term, squared
says someone still trying to make the special case of Exponents and Brackets apply to a Factorised Term when it doesn't. 😂 I'll take that as your admission of being wrong about a(b+c)=ax(b+c) then. Thanks for playing
Only if you had defined it as such to begin with, otherwise the Brackets Exponents rule doesn't apply if you started out with 2(8)², which is different to 2(8²) and 2(ab)²
..and there's still no exponent in a(b+c) anyway, Mr. False Equivalence