Question

X^2-y^2/x+y-x^3-y^3/x^2+xy+y^2

Answer 1

Все через формулы сокращённого умножения

Answer 2

To simplify the given expression, let's factor the numerator and denominator separately and then simplify:

Expression: (x^2 - y^2) / (x + y) - (x^3 - y^3) / (x^2 + xy + y^2)

Step 1: Factor the numerator and denominator in the first term (x^2 - y^2) / (x + y): (x^2 - y^2) is the difference of squares and can be factored as (x + y)(x - y).

So the first term becomes: [(x + y)(x - y)] / (x + y)

Step 2: Factor the numerator and denominator in the second term (x^3 - y^3) / (x^2 + xy + y^2): (x^3 - y^3) is the difference of cubes and can be factored as (x - y)(x^2 + xy + y^2).

So the second term becomes: [(x - y)(x^2 + xy + y^2)] / (x^2 + xy + y^2)

Step 3: Now, we can simplify the expression: [(x + y)(x - y)] / (x + y) - [(x - y)(x^2 + xy + y^2)] / (x^2 + xy + y^2)

Since the denominators are the same, we can combine the fractions: = [(x + y)(x - y) - (x - y)(x^2 + xy + y^2)] / (x^2 + xy + y^2)

Now, let's factor out (x - y) from the numerator: = (x - y)[(x + y) - (x^2 + xy + y^2)] / (x^2 + xy + y^2)

Next, combine like terms in the brackets: = (x - y)[x + y - x^2 - xy - y^2] / (x^2 + xy + y^2)

Now, factor out -1 from the second and third terms in the brackets: = (x - y)[x + y - (x^2 + xy + y^2)] / (x^2 + xy + y^2)

Notice that the expression inside the brackets is -(x^2 + xy + y^2) + (x + y). We can rewrite it as (x + y) - (x^2 + xy + y^2):

= (x - y)[(x + y) - (x^2 + xy + y^2)] / (x^2 + xy + y^2) = (x - y)[(x + y) - (x + y)^2] / (x^2 + xy + y^2)

Since (x + y)^2 = x^2 + 2xy + y^2, we can continue simplifying: = (x - y)[(x + y) - (x^2 + 2xy + y^2)] / (x^2 + xy + y^2) = (x - y)(-x^2 - xy - y^2) / (x^2 + xy + y^2)

Finally, we can factor out -1 from the numerator: = -(x - y)(x^2 + xy + y^2) / (x^2 + xy + y^2)

And now the (x^2 + xy + y^2) terms cancel out, leaving us with the simplified expression: = -(x - y)

So the simplified expression is -(x - y).