Question

1) factor of x³-1/x³​

Answer 1

Answer:

$x^{3} -\frac{1}{x^{3}} = \frac{1}{x^{3}} [(x^{2} -1)(x^{4}+x^{2} +1)]$

Step-by-step explanation:

• Given expression is $x^{3}-\frac{1}{x^{3}}$

• In order to solve this, first we have to do cross multiplication to make denominators equal, we get

       $x^{3}-\frac{1}{x^{3}}= \frac{x^{6}-1}{x^{3}}$  

                  $= \frac{1}{x^{3}}(x^{6} -1)$

                  $= \frac{1}{x^{3}} [(x^{2})^{3} -1^{3} ]$        

• We know the identity,  

         $a^{3} -a^{3} =(a-b)(a^{2} +ab+b^{2} )$  

• By applying this formula in above equation, we get

          $= \frac{1}{x^{3} } [(x^{2} -1)(x^{4} +x^{2} +1)]$

Answer 2

Given,

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

Here, we have to factorize the given expression,

The given expression same like $a^{3}-b^{3}$ identity form

Formula,

$a^{3}-b^{3}= (a-b)(a^{2}+ab+b^{2} )$

From the given,

$a= x$, $b= \frac{1}{x}$

Now substitute the values in above formula, we get

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

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

$= (x-\frac{1}{x} )(\frac{x^{4}+x^{2} +1}{x^{2} } )$

$(x-\frac{1}{x} )(\frac{x^{4}+x^{2} +1}{x^{2} } )=0$

Therefore, the factors of $x^{3}-\frac{1}{x^{3} }$ is $(x-\frac{1}{x} )(\frac{x^{4}+x^{2} +1}{x^{2} } )$.