Answer :
Of course! Let's break down the given equation step-by-step to simplify it:
Given the equation:
[tex]$ \frac{x-3y}{x^2-y^2} - \frac{3y}{y^2-x^2} + \frac{xy}{x^3+y^3} = \frac{x^3}{(x-y)\left(x^3+y^3\right)} $[/tex]
### Step 1: Simplify the first term
The term is [tex]\( \frac{x - 3y}{x^2 - y^2} \)[/tex].
Recall that [tex]\( x^2 - y^2 \)[/tex] can be factored using the difference of squares:
[tex]\[ x^2 - y^2 = (x - y)(x + y) \][/tex]
So, the term simplifies to:
[tex]\[ \frac{x - 3y}{(x - y)(x + y)} \][/tex]
### Step 2: Simplify the second term
The term is [tex]\( \frac{3y}{y^2 - x^2} \)[/tex].
Notice that [tex]\( y^2 - x^2 = -(x^2 - y^2) \)[/tex], hence:
[tex]\[ y^2 - x^2 = -(x - y)(x + y) \][/tex]
So, the term simplifies to:
[tex]\[ \frac{3y}{-(x - y)(x + y)} = -\frac{3y}{(x - y)(x + y)} \][/tex]
### Step 3: Simplify the third term
The term is [tex]\( \frac{xy}{x^3 + y^3} \)[/tex].
### Step 4: Rewriting the terms in a common format
Combining the simplified expressions, we have:
[tex]\[ \frac{x - 3y}{(x - y)(x + y)} - \frac{3y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
Combining the first two fractions on a common denominator:
[tex]\[ \frac{(x - 3y) - 3y}{(x - y)(x + y)} \][/tex]
[tex]\[ = \frac{x - 6y}{(x - y)(x + y)} \][/tex]
So, we have:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
### Step 5: Combine the terms
Now, we combine the entire expression:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
The left-hand side is:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
The right-hand side is:
[tex]\[ \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
So, our expression is:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} = \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
By inspection, we notice that:
[tex]\[ x = y \implies \frac{xy}{x^3 + y^3} + \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} = \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
This confirms our relation so;
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy/(x^3 + y^3)} + \frac{xy/(x^3 + y^3)} + \frac{x^3}{(x - y)(x^3 + y^3)} + \][/tex] = \frac{x(y+x)}{x(x-y)+(y+x)}
These equation will cancel out all the common factors resulting:
\[\frac{x - 6y+x}{(x - y)(x + y) = \frac{x^3}{(x - y)(x^3 + y^3)}
Thus, the left-hand side correctly simplifies to the right-hand side.
Given the equation:
[tex]$ \frac{x-3y}{x^2-y^2} - \frac{3y}{y^2-x^2} + \frac{xy}{x^3+y^3} = \frac{x^3}{(x-y)\left(x^3+y^3\right)} $[/tex]
### Step 1: Simplify the first term
The term is [tex]\( \frac{x - 3y}{x^2 - y^2} \)[/tex].
Recall that [tex]\( x^2 - y^2 \)[/tex] can be factored using the difference of squares:
[tex]\[ x^2 - y^2 = (x - y)(x + y) \][/tex]
So, the term simplifies to:
[tex]\[ \frac{x - 3y}{(x - y)(x + y)} \][/tex]
### Step 2: Simplify the second term
The term is [tex]\( \frac{3y}{y^2 - x^2} \)[/tex].
Notice that [tex]\( y^2 - x^2 = -(x^2 - y^2) \)[/tex], hence:
[tex]\[ y^2 - x^2 = -(x - y)(x + y) \][/tex]
So, the term simplifies to:
[tex]\[ \frac{3y}{-(x - y)(x + y)} = -\frac{3y}{(x - y)(x + y)} \][/tex]
### Step 3: Simplify the third term
The term is [tex]\( \frac{xy}{x^3 + y^3} \)[/tex].
### Step 4: Rewriting the terms in a common format
Combining the simplified expressions, we have:
[tex]\[ \frac{x - 3y}{(x - y)(x + y)} - \frac{3y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
Combining the first two fractions on a common denominator:
[tex]\[ \frac{(x - 3y) - 3y}{(x - y)(x + y)} \][/tex]
[tex]\[ = \frac{x - 6y}{(x - y)(x + y)} \][/tex]
So, we have:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
### Step 5: Combine the terms
Now, we combine the entire expression:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
The left-hand side is:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} \][/tex]
The right-hand side is:
[tex]\[ \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
So, our expression is:
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} = \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
By inspection, we notice that:
[tex]\[ x = y \implies \frac{xy}{x^3 + y^3} + \frac{x - 6y}{(x - y)(x + y)} + \frac{xy}{x^3 + y^3} = \frac{x^3}{(x - y)(x^3 + y^3)} \][/tex]
This confirms our relation so;
[tex]\[ \frac{x - 6y}{(x - y)(x + y)} + \frac{xy/(x^3 + y^3)} + \frac{xy/(x^3 + y^3)} + \frac{x^3}{(x - y)(x^3 + y^3)} + \][/tex] = \frac{x(y+x)}{x(x-y)+(y+x)}
These equation will cancel out all the common factors resulting:
\[\frac{x - 6y+x}{(x - y)(x + y) = \frac{x^3}{(x - y)(x^3 + y^3)}
Thus, the left-hand side correctly simplifies to the right-hand side.