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Factorize-2-1-x-3-x-3-Stepwise-process-is-required-

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Question Number 2134 by Rasheed Soomro last updated on 04/Nov/15
Factorize −2+(1/x^3 )+x^3 (Stepwise process is required)
Factorize−2+1x3+x3(Stepwiseprocessisrequired)
Answered by sudhanshur last updated on 04/Nov/15
−2+(1/x^3 )+x^3 =((−2x^3 +1+x^6 )/x^3 )=(((x^3 −1)^2 )/x^3 ) =(((x−1)^2 (x^2 +x+1)^2 )/x^3 )
−2+1x3+x3=−2x3+1+x6x3=(x3−1)2x3=(x−1)2(x2+x+1)2x3
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: ((1/x))^3 (x−1)^2 (x^2 +x+1)^2 .
Anewapproach,Ididn′tthinkof.Infactorformmoresuitablycanbewritten:(1x)3(x−1)2(x2+x+1)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 ))
−2+1x3+x3=1+1x3+x3−3a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)=(1)3+(1x)3+(x)3−3(1)(1x)(x)=(1+1x+x)[(1)2+(1x)2+(x)2−(1)(1x)−(1x)(x)−(x)(1)]=(x+1x+1)(x2+1x2+1×−x−1x−1×)=(x+1x+1)(x2+1x2−x−1x)Factorsofx2+1x2−x−1xsuggestedbysudhanshur=(x+1x+1)(x2−x+1x2−1x)=(x+1x+1)[x(x−1)−x−1x2]=(x+1x+1)(x−1)(x−1x2)
Commented by sudhanshur last updated on 04/Nov/15
(x^2 +(1/x^2 )−x−(1/x))=x^2 −x−((x−1)/x^2 )=(x−1)(x−(1/x^2 ))
(x2+1x2−x−1x)=x2−x−x−1x2=(x−1)(x−1x2)
Commented by Rasheed Soomro last updated on 05/Nov/15
Good!
Good!

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