# How to verify an identity[1-3x^2/(1-x^2)]*(1-x)=(1-2x)*[1+x/(x+1)]

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### 2 Answers

We have to verify [1-3x^2/(1-x^2)]*(1-x)=(1-2x)*[1+x/(x+1)]

If [1-3x^2/(1-x^2)]*(1-x)=(1-2x)*[1+x/(x+1)]

=> [1-3x^2/(1-x^2)]*(1-x) - (1-2x)*[1+x/(x+1)] = 0

[1-3x^2/(1-x^2)]*(1-x) - (1-2x)*[1+x/(x+1)]

=> (1 - x^2 - 3x^2)*(1 - x)/(1 - x^2) - (1 - 2x)(2x + 1)/(x + 1)

=> (1 - x^2 - 3x^2)/(1 + x) - (1 - 4x^2)/(x + 1)

=> (1 - 4x^2 - 1 + 4x^2)/(1 + x)

=> 0

As we get 0 , the identity is proved.

We'll manipulate what's inside the 1st pair of brackets:

1-3x^2/(1-x^2) = (1 - x^2 - 3x^2)/(1-x)(1+x) = (1-4x^2)/(1-x^2)

We'll multiply (1-4x^2)/(1-x^2) by (1-x)

(1-4x^2)(1-x)/(1-x^2)

We'll re-write the denominator as a product, since it is a difference of 2 squares:

(1-4x^2)(1-x)/(1-x)(1+x)

We'll simplify and we'll get:

(1-4x^2)/(1+x)

We'll manipulate what's inside the 2nd pair of brackets:

x/(x+1) + 1

We'll multiply by (x+1) the number 1:

[x + 1*(x+1)]/(x+1) = (x + x + 1)/(x+1) = (2x+1)/(x+1)

We'll multiply (2x+1)/(x+1) by (1-2x)

(2x+1)(1-2x)/(x+1)

We notice that the product from numerator yields a difference of 2 squares:

(2x+1)(1-2x)/(x+1) = (1-4x^2)/(1+x) = LHS

Since we've get LHS=RHS, the given identity is verified.

All you have to do is to manipulate each side, at a time.