From the factoring, it is possible to find the following answers:
Letter A: [tex]\frac{x\left(x+2\right)}{x-3}[/tex]
Letter B: [tex]-\frac{2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}[/tex]
Letter C: [tex]x^2+y^2[/tex]
Factoring
In math, factoring or factorization is used to write an algebraic expression in factors. There are some rules for factorization. One of them is a factor out a common term for example: x²-x= x(x-1), where x is a common term.
Here you should factor the given expression.
[tex]\frac{x^3-4x}{x^2-5x+6} \\ \\ \frac{x\left(x+2\right)\left(x-2\right)}{x^2-5x+6}\\ \\ \frac{x\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ \\ \frac{x\left(x+2\right)}{x-3}\\ \\ Then,\\ \\ \frac{x^3-4x}{x^2-5x+6} =\frac{x\left(x+2\right)}{x-3}[/tex]
Firstly, you should replace the variables A, B and C for the given expressions.
[tex]\frac{1}{1-x+x^2}-\frac{1}{1+x+x^2}-\frac{2x}{1+x^2}[/tex]
After that, you should find the least common multiple.
[tex]\frac{\left(x^2+1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{\left(x^2+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{2x\left(x^2-x+1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}[/tex]
Finally, you can simplify the expression
[tex]\frac{\left(x^2+1\right)\left(x^2+x+1\right)-\left(x^2+1\right)\left(x^2-x+1\right)-2x\left(x^2-x+1\right)\\ \\ \left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\\ \\ \frac{-2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}= -\frac{2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}[/tex]
Firstly, you should replace the variables P, Q and R for the given expressions.
[tex]\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}\div \frac{x+y}{x^2+y^2+xy}[/tex]
Rewriting
[tex]\frac{\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}}{\frac{x+y}{x^2+y^2+xy}}=\frac{\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}\left(x^2+y^2+xy\right)}{x+y}\\ \\[/tex]
[tex]\frac{\left(x^2+y^2\right)\left(x+y\right)}{x+y}=x^2+y^2[/tex]
Read more about the factoring here:
brainly.com/question/11579257
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