this post was submitted on 27 Nov 2025
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Juxtaposition is key to the bullshit you made up, you infuriating sieve. You made a hundred comments in this thread about how 2*(8)^2^ is different from 2(8)^2^. Here is a Maths textbook saying, you're fucking wrong.
Here's another: First Steps In Algebra, Wentworth 1904, on page 143 (as in the Gutenberg PDF), in exercise 54, question 9 reads (x-a)(2x-a)=2(x-b)^2^. The answer on page 247 is x=(2b^2^-a^2^)/(4b-3a). If a=1, b=0, the question and answer get 1/3, and the bullshit you've made up does not.
You have harassed a dozen people specifically to insist that 6(ab)^2^ does not equal 6a^2^b^2^. You've sassed me specifically to say a variable can be zero, so 6(a+b) can be 6(a+0) can just be 6(a). There is no out for you. This is what you've been saying, and you're just fucking wrong, about algebra, for children.
Terms/Products is mathematical fact, as is The Distributive Law. Maths textbooks never use the word "juxtaposition".
That's right. 1/2(8)²=1/256, 1/2x8²=32, same difference as 8/2(1+3)=1 but 8/2x(1+3)=16
Nope! It doesn't say that 1/a(b+c)=1/ax(b+c). You're making a false equivalence argument
Question about solving an equation and not about solving an expression. False equivalence again.
Nope! I have never said that, which is why you're unable to quote me saying that. I said 6(a+b)² doesn't equal 6x(a+b)², same difference as 8/2(1+3)=1 but 8/2x(1+3)=16
That's right
Got no idea what you're talking about
Yes
No, you've come up with nothing other than False Equivalence arguments. You're taking an equation with exponents and no division, and trying to say the same rules apply to an expression with division and no exponents, even though we know that exponent rule is a special case anyway, even if there was an exponent in the expression, which there isn't. 🙄
For teenagers, who are taught The Distributive Law in Year 7
...
That's you saying it. You are unambiguously saying a(b)^c^ somehow means (ab)^c^=a^c^b^c^ instead of ab^c^, except when you try to nuh-uh at anyone pointing out that's what you said. Where the fuck did 256 come from if that's not exactly what you're doing?
You're allegedly an algebra teacher, snipping about terms I am quoting from a textbook you posted, and you wanna pretend 2(x-b)^2^ isn't precisely what you insist you're talking about? Fine, here's yet another example:
A First Book In Algebra, Boyden 1895, on page 47 (49 in the Gutenberg PDF), in exercise 24, question 18 reads, divide 15(a-b)^3^x^2^ by 3(a-b)x. The answer on page 141 of the PDF is 5(a-b)^2^x. For a=2, b=1, the question and answer get 5x, while the bullshit you've made up gets 375x.
Show me any book where the equations agree with you. Not words, not acronyms - an answer key, or a worked example. Show me one time that published math has said x(b+c)^n^ gets an x^n^ term. I've posted four examples to the contrary and all you've got is pretending not to see x(b+c)^n^ right fuckin' there in each one.
No it isn't! 😂 Spot the difference 1/2(8)²=1/256 vs.
Nope. Never said that either 🙄
Because that isn't what I said. See previous point 😂
From 2(8)², which isn't the same thing as 2(ab)² 🙄 The thing you want it to mean is 2(8²)
Because you're on a completely different page and making False Equivalence arguments.
No idea what you're talking about, again. 2(x-b)2 is most certainly different to 2(xb)2, no pretense needed. you're sure hung up on making these False Equivalence arguments.
Easy. You could've started with that and saved all this trouble. (you also would've found this if you'd bothered to read my thread that I linked to)...
Thus, x(x-1) is a single term which is entirely in the denominator, consistent with what is taught in the early chapters of the book, which I have posted screenshots of several times.
You've posted 4 False Equivalence arguments 🙄 If you don't understand what that means, it means proving that ab=axb does not prove that 1/ab=1/axb. In the former there is multiplication only, in the latter there is Division, hence False Equivalence in trying to say what applies to Multiplication also applies to Division
Pointing out that you're making a False Equivalence argument. You're taking examples where the special Exponent rule of Brackets applies, and trying to say that applies to expressions with no Exponents. It doesn't. 🙄 The Distributive Law always applies. The special exponent rule with Brackets only applies in certain circumstances. I already said this several posts back, and you're pretending to not know it's a special case, and make a False Equivalence argument to an expression that doesn't even have any exponents in it 😂
a=8, b=1, it's the same thing.
False equivalence is you arguing about brackets and exponents by pointing to equations without exponents.
This entire thing is about your lone-fool campaign to insist 2(8)^2^ doesn't mean 2*8^2^, despite multiple textbook examples that only work because a(b)^c^ is a*b^c^ and not a^c^b^c^.
I found four examples, across two centuries, of your certain circumstances: addition in brackets, factor without multiply symbol, exponent on the bracket. You can't pivot to pretending this is a division syntax issue, when you've explicitly said 2(8)^2^ is (2*8)^2^. Do you have a single example that matches that, or are you just full of shit?
No it isn't! 😂 8 is a single numerical factor. ab is a Product of 2 algebraic factors.
Nope. I was talking about 1/a(b+c) the whole time, as the reason the Distributive Law exists, until you lot decided to drag exponents into it in a False Equivalence argument. I even posted a textbook that showed more than a century ago they were still writing the first set of Brackets. i.e. 1/(a)(b+c). i.e. It's the FOIL rule, (a+b)(c+d)=(ac+ad+bc+bd) where b=0, and these days we don't write (a)(b+c) anymore, we just write a(b+c), which is already a single Term, same as (a+b)(c+d) is a single term, thus doesn't need the brackets around the a to show it's a single Term.
3(x-y) is a single term...
Hilarious that all Maths textbooks, Maths teachers, and most calculators agree with me then, isn't it 😂
Again, you lot were the ones who dragged exponents into it in a False Equivalence argument to 1/a(b+c)
None of which relate to the actual original argument about 1/a(b+c)=1/(ab+ac) and not (b+c)/a
I'm not pivoting, that was the original argument. 😂 The most popular memes are 8/2(2+2) and 6/2(1+2), and in this case they removed the Division to throw a curve-ball in there (note the people who failed to notice the difference initially). We know a(b+c)=(ab+ac), because it has to work when it follows a Division, 1/a(b+c)=1/(ab+ac). It's the same reason that 1/a²=1/(axa), and not 1/axa=a/a=1. It's the reason for the brackets in (ab+ac) and (axa), hence why it's done in the Brackets step (not the MULTIPLY step). It's you lot trying to pivot to arguments about exponents, because you are desperately trying to separate the a from a(b+c), so that it can be ax(b+c), and thus fall to the Multiply step instead of the Brackets step, but you cannot find any textbooks that say a(b+c)=ax(b+c) - they all say a(b+c)=(ab+ac) - so you're trying this desperate False Equivalence argument to separate the a by dragging exponents into it and invoking the special Brackets rule which only applies in certain circumstances, none of which apply to a(b+c) 😂
That's right
says the person making False Equivalence arguments. 🙄 Let me know when you find a textbook that says a(b+c)=ax(b+c), otherwise I'll take that as an admission of being wrong that you keep avoiding the actual original point that a(b+c) is a single Term, as per Maths textbooks
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