Question Number 2134 by Rasheed Soomro last updated on 04/Nov/15
$${Factorize} \\ $$$$−\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}^{\mathrm{3}} \\ $$$$\left({Stepwise}\:{process}\:{is}\:{required}\right) \\ $$
Answered by sudhanshur last updated on 04/Nov/15
$$−\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}^{\mathrm{3}} \\ $$$$=\frac{−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}+{x}^{\mathrm{6}} }{{x}^{\mathrm{3}} }=\frac{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{3}} } \\ $$$$=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{3}} } \\ $$
Commented by Rasheed Soomro last updated on 04/Nov/15
$${A}\:{new}\:{approach},\:{I}\:{didn}'{t}\:{think}\:{of}.{In}\:{factor}\:{form}\:{more} \\ $$$${suitably}\:{can}\:{be}\:{written}: \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:. \\ $$
Answered by Rasheed Soomro last updated on 05/Nov/15
![−2+(1/x^3 )+x^3 =1+(1/x^3 )+x^3 −3 a^3 +b^3 +c^3 −3abc =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) =(1)^3 +((1/x))^3 +(x)^3 −3(1)((1/x))(x) =(1+(1/x)+x)[(1)^2 +((1/x))^2 +(x)^2 −(1)((1/x))−((1/x))(x)−(x)(1)] =(x+(1/x)+1)(x^2 +(1/x^2 )+1^(×) −x−(1/x)−1^(×) ) =(x+(1/x)+1)(x^2 +(1/x^2 )−x−(1/x)) Factors of x^2 +(1/x^2 )−x−(1/x) suggested by sudhanshur =(x+(1/x)+1)(x^2 −x+(1/x^2 )−(1/x)) =(x+(1/x)+1)[x(x−1)−((x−1)/x^2 )] =(x+(1/x)+1)(x−1)(x−(1/x^2 ))](https://www.tinkutara.com/question/Q2147.png)
$$−\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}^{\mathrm{3}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}^{\mathrm{3}} −\mathrm{3} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\left({x}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{1}\right)\left(\frac{\mathrm{1}}{{x}}\right)\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}+{x}\right)\left[\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\left({x}\right)^{\mathrm{2}} −\left(\mathrm{1}\right)\left(\frac{\mathrm{1}}{{x}}\right)−\left(\frac{\mathrm{1}}{{x}}\right)\left({x}\right)−\left({x}\right)\left(\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\overset{×} {\mathrm{1}}−{x}−\frac{\mathrm{1}}{{x}}−\overset{×} {\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$${Factors}\:{of}\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}−\frac{\mathrm{1}}{{x}}\:{suggested}\:{by}\:{sudhanshur} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left[{x}\left({x}−\mathrm{1}\right)−\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} }\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by sudhanshur last updated on 04/Nov/15
$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}−\frac{\mathrm{1}}{{x}}\right)={x}^{\mathrm{2}} −{x}−\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} }=\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$
Commented by Rasheed Soomro last updated on 05/Nov/15
$${Good}! \\ $$