Factorize a3 +b3 +c3 −3abc

Solution: The expression we want to factorize is a 3 +b3 +c3 −3abc. To simplify it, we'll use the identity: (x + y) 3 = x3 + y3+ 3xy(x + y) · In our case, we can see that: So, we have: a 3 + b3 + c3 - 3abc = (a + b)3 + c3- 3abc · Now, let's factorize the expression (a + b) 3 + c. Now, let's factorize the expression (a + b) 3 + c3 - 3abc using the identity: (x + y) 3 = x3 + y3 + 3xy(x + y) · In our case: So, we can rewrite the expression as: (a + b) 3 + c3 - 3abc = (a + b + c)((a + b)2 - (a + b)c + c2) - 3abc ...

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Algebraic identities P 2 (a,b)=a 3 +b 3 =(a+b)(a 2 +b 2 −ab) P 3 (a,b,c)=a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−ac−bc) P 4 (a,b,c,d)=a 3 +b 3 +c 3 +d 3 −3abc−3abd...

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(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)−abc I did it a 2 b+a 2 c+ab 2 +cb 2 +bc 2 +ac 2 +2abc=a 2 (b+c)+bc(b+c)+a(b 2 +c 2 +2bc) =a 2 (b+c)+bc(b+c)+a(b+c) 2 =(b+c)(a 2 +bc+ab+ac) =(b+c)[a(a+b)...

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S → iEtS | iEtSeS | a E → b left factoring S -> iEtSS' | a S' -> ε | eS E -> b A → aAB / aBc / aAc left factoring A -> aA' A' -> AD | Bc D -> B | c S → bSSaaS / bSSaSb / bSb / a left factoring S → bSS' / a S...

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