Question

Express the expression $\frac{1}{1 - x} - \frac{1}{1 + x} - \frac{ x^{3} }{1 - x} + \frac{ x^{2} }{1 + x}$ in lowest terms .

Answer 1

The answer is explained below.


Hope this helps!

Answer 2

☯ Given:

$\frac{1}{1-x} -\frac{1}{1+x} -\frac{x^{3} }{1-x} +\frac{x^{2} }{1+x}$

☯ To Find:

Lowest term of given Expression

☯ Solution:

First, consider the first and 3rd term of the expression and simplify it.

⇒   $\frac{1-x^{3} }{1-x}$  = $\frac{(1-x)(1+x^{2} +x)}{1-x}$   (By the formula of a³ - b³)

Now, considering 2nd and 4th term;

⇒  $\frac{x^{2} -1}{x+1}$  =  $\frac{(x+1)(x-1)}{x+1}$    (By the formula a² - b²)

Now, adding these two, we get;

$\frac{(1-x)(1+x^{2} +x)}{1-x}$  +  $\frac{(x+1)(x-1)}{x+1}$   =    $\frac{(1+x)(1-x)(x^{2} +x+1) + (1-x)(1+x)(x-1)}{1-x^{2} }$

⇒ $(1+x)(1-x)[\frac{x^{2} +x+1)+(x-1)}{1-x^{2} } ]$

=  $\frac{1-x^{2} }{1-x^{2} } [(x^{2} +x+1)+(x-1)]$

= $(x^{2} +x+1+x-1)$

= $x^{2} +2x$

$x(x+2)$

☯ Final Answer:

$x(x+2)$