B Algebra Help Needed: Understanding Algebraic Manipulations and Techniques

[Jehannum]

I'm reading a textbook that does a couple of algebraic manipulations in one step (in order to get the expressions ready for the binomial theorem). I'm unfamiliar with this step. Could anyone expand it into several steps? It's using some kind of algebraic standard technique or result that I don't know.

Here's the first one:

(x + dx) ^ -2 becomes x ^ -2 (1 + dx / x) ^ -2​

The second one is:

(x + dx) ^ 0.5 becomes x ^ 0.5 (1 + dx / x) ^ 0.5​

Am I correct in inferring that there's a general rule:

(a + b) ^ c becomes a ^ c (1 + b / a) ^ c​

If so, is there a name for this rule, and are there any conditions on the values of a, b and c for it to work?

 

[fresh_42]

The first rule to be applied is $(x\cdot y)^p=x^p \cdot y^p$ where the associativity ($x\cdot (y \cdot z) = (x \cdot y) \cdot z$) and commutativity ($x \cdot y = y \cdot x$) of multiplication is used and which gives $a^c \cdot (1+\frac{b}{a})^c=(a \cdot (1+\frac{b}{a}))^c$. Next we use the distributive law $x \cdot (y+z)=x \cdot z+y\cdot z$ to get $(a \cdot (1+\frac{b}{a}))^c=(a \cdot 1 +a \cdot \frac{b}{a})^c=(a+b)^c$ where again associativity of multiplication is used. The only condition that has to be valid is $a \neq 0$ for otherwise the division wouldn't be defined.

 

[PeroK]

I'm reading a textbook that does a couple of algebraic manipulations in one step (in order to get the expressions ready for the binomial theorem). I'm unfamiliar with this step. Could anyone expand it into several steps? It's using some kind of algebraic standard technique or result that I don't know.

Am I correct in inferring that there's a general rule:

(a + b) ^ c becomes a ^ c (1 + b / a) ^ c​

If so, is there a name for this rule, and are there any conditions on the values of a, b and c for it to work?

That's a basic property of exponents.

https://www.mathsisfun.com/algebra/exponent-laws.html

 

[Jehannum]

The first rule to be applied is $(x\cdot y)^p=x^p \cdot y^p$ where the associativity ($x\cdot (y \cdot z) = (x \cdot y) \cdot z$) and commutativity ($x \cdot y = y \cdot x$) of multiplication is used and which gives $a^c \cdot (1+\frac{b}{a})^c=(a \cdot (1+\frac{b}{a}))^c$. Next we use the distributive law $x \cdot (y+z)=x \cdot z+y\cdot z$ to get $(a \cdot (1+\frac{b}{a}))^c=(a \cdot 1 +a \cdot \frac{b}{a})^c=(a+b)^c$ where again associativity of multiplication is used. The only condition that has to be valid is $a \neq 0$ for otherwise the division wouldn't be defined.

Thank you. That's clear now.