[tex]\frac{a-b}{b}*\left(\frac{b}{b-a}+\frac{b}{a}\right)[/tex]
First, let's simplify the expression inside the parentheses:
[tex]\frac{b}{b-a} + \frac{b}{a}[/tex]
This can be rewritten as:
[tex]\frac{ba}{a(b-a)} + \frac{b(b-a)}{a(b-a)}[/tex]
Combining these fractions, we get:
[tex]\frac{ba + b(b-a)}{a(b-a)} = \frac{ab + bb - ba}{a(b-a)} = \frac{b^2}{a(b-a)}[/tex]
Now, let's substitute this back into the original expression:
[tex]\frac{a-b}{b}\left(\frac{b^2}{a(b-a)}\right) = \frac{a-b}{b}\frac{b^2}{a(b-a)}[/tex]
Multiplying this out gives us:
[tex]\frac{(a-b)b^2}{a(b-a)b} = \frac{ab^2 - b^3}{ab(b-a)}[/tex]
So, the final simplified expression is [tex]\frac{ab^2 - b^3}{ab*(b-a)}[/tex]
[tex]\frac{a-b}{b}*\left(\frac{b}{b-a}+\frac{b}{a}\right)[/tex]
First, let's simplify the expression inside the parentheses:
[tex]\frac{b}{b-a} + \frac{b}{a}[/tex]
This can be rewritten as:
[tex]\frac{ba}{a(b-a)} + \frac{b(b-a)}{a(b-a)}[/tex]
Combining these fractions, we get:
[tex]\frac{ba + b(b-a)}{a(b-a)} = \frac{ab + bb - ba}{a(b-a)} = \frac{b^2}{a(b-a)}[/tex]
Now, let's substitute this back into the original expression:
[tex]\frac{a-b}{b}\left(\frac{b^2}{a(b-a)}\right) = \frac{a-b}{b}\frac{b^2}{a(b-a)}[/tex]
Multiplying this out gives us:
[tex]\frac{(a-b)b^2}{a(b-a)b} = \frac{ab^2 - b^3}{ab(b-a)}[/tex]
So, the final simplified expression is [tex]\frac{ab^2 - b^3}{ab*(b-a)}[/tex]