Question

x3+x2-1/x2+1/x3 factorise​

Answer 1

Answer:

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

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

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

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

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

Answer 2

The  factor of $x^{3} +x^{2} -\frac{1}{x^{2} } +\frac{1}{x^{3} }$ is $(x+\frac{1}{x} )(x^{2} +\frac{1}{x^{2} } +x-\frac{1}{x} -1 })$.

Step-by-step explanation:

Given:

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

To Find:

The  factor of $x^{3} +x^{2} -\frac{1}{x^{2} } +\frac{1}{x^{3} }$.

Solution:

As given,$x^{3} +x^{2} -\frac{1}{x^{2} } +\frac{1}{x^{3} }$.

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

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

Using formula $p^{3} +q^{3} =(p+q)(p^{2}-pq+q^{2} )$ and $p^{2} -q^{2} =(p+q)(p-q)$.

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

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

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

Using formula $(p+q)^{2} =p^{2} +q^{2} +2pq$.

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

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

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

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

Thus,the  factor of $x^{3} +x^{2} -\frac{1}{x^{2} } +\frac{1}{x^{3} }$ is $(x+\frac{1}{x} )(x^{2} +\frac{1}{x^{2} } +x-\frac{1}{x} -1 })$.

#SPJ2