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Question Number 180894 by depressiveshrek last updated on 18/Nov/22

x^3 +x=1 x^8 +3x^3 =?

$${x}^{\mathrm{3}} +{x}=\mathrm{1} \\ $$$${x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} =? \\ $$

Commented by Rasheed.Sindhi last updated on 19/Nov/22

The solutions/answers (complete process) should be posted as “Answer” not as comment (kindly).

$$\mathcal{T}{he}\:{solutions}/{answers}\:\left({complete}\right. \\ $$$$\left.{process}\right)\:{should}\:{be}\:{posted}\:{as}\:``\mathrm{Answer}'' \\ $$$${not}\:{as}\:{comment}\:\left({kindly}\right). \\ $$

Commented by Gamil last updated on 19/Nov/22

Answered by Frix last updated on 18/Nov/22

x^8 +3x^3 −a=0 ((x^8 +3x^3 −a)/(x^3 +x−1))=x^5 −x^3 +x^2 +x+1−((1−a)/(x^3 +x−1)) ⇒ a=1

$${x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} −{a}=\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} −{a}}{{x}^{\mathrm{3}} +{x}−\mathrm{1}}={x}^{\mathrm{5}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}−\frac{\mathrm{1}−{a}}{{x}^{\mathrm{3}} +{x}−\mathrm{1}} \\ $$$$\Rightarrow\:{a}=\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 19/Nov/22

x^3 +x−1=0 How have you divided with 0 sir?

$${x}^{\mathrm{3}} +{x}−\mathrm{1}=\mathrm{0}\: \\ $$$$\mathcal{H}{ow}\:{have}\:{you}\:{divided}\:{with}\:\mathrm{0}\:{sir}? \\ $$

Commented by Frix last updated on 19/Nov/22

You can do this because if two polynomes have common factors it will just eliminate these common factors. The function f(x)=(((x^2 −1)(x+3))/((x^2 +2)(x+3))) is not defined at x=−3 but if we add the lim_(x→−3) f(x) (and in most cases it makes sense to add it) it becomes the same as g(x)=((x^2 −1)/(x^2 +2)).

$$\mathrm{You}\:\mathrm{can}\:\mathrm{do}\:\mathrm{this}\:\mathrm{because}\:\mathrm{if}\:\mathrm{two}\:\mathrm{polynomes} \\ $$$$\mathrm{have}\:\mathrm{common}\:\mathrm{factors}\:\mathrm{it}\:\mathrm{will}\:\mathrm{just}\:\mathrm{eliminate} \\ $$$$\mathrm{these}\:\mathrm{common}\:\mathrm{factors}. \\ $$$$\mathrm{The}\:\mathrm{function}\:{f}\left({x}\right)=\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}+\mathrm{3}\right)}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}+\mathrm{3}\right)}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{defined}\:\mathrm{at}\:{x}=−\mathrm{3}\:\mathrm{but}\:\mathrm{if}\:\mathrm{we}\:\mathrm{add}\:\mathrm{the}\:\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}{f}\left({x}\right) \\ $$$$\left(\mathrm{and}\:\mathrm{in}\:\mathrm{most}\:\mathrm{cases}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{to}\:\mathrm{add}\:\mathrm{it}\right) \\ $$$$\mathrm{it}\:\mathrm{becomes}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:{g}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}}. \\ $$

Commented by Rasheed.Sindhi last updated on 19/Nov/22

ThanX sir!

$$\mathcal{T}{han}\mathcal{X}\:\boldsymbol{{sir}}! \\ $$

Answered by Rasheed.Sindhi last updated on 19/Nov/22

x^3 +x=1; x^8 +3x^3 =? x^3 =1−x ▶x^8 +3x^3 =(x^3 )^2 ∙x^2 +3x^3 =x^2 (1−x)^2 +3(1−x) =x^2 (1−2x+x^2 )+3−3x =x^2 −2x^3 +x^3 ∙x+3−3x =x^2 −2(1−x)+x(1−x)+3−3x =x^2 −2+2x+x−x^2 +3−3x =1

$${x}^{\mathrm{3}} +{x}=\mathrm{1};\:{x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} =? \\ $$$${x}^{\mathrm{3}} =\mathrm{1}−{x} \\ $$$$\:\blacktriangleright{x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} =\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \centerdot{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right)+\mathrm{3}−\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{3}} \centerdot{x}+\mathrm{3}−\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{x}\right)+{x}\left(\mathrm{1}−{x}\right)+\mathrm{3}−\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:=\cancel{{x}^{\mathrm{2}} }−\mathrm{2}+\mathrm{2}{x}+{x}−\cancel{{x}^{\mathrm{2}} }+\mathrm{3}−\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Answered by mr W last updated on 19/Nov/22

x^8 +3x^3 =x^8 +x^6 −x^6 −x^4 +x^4 +x^2 −x^2 +3x^3 =x^5 (x^3 +x)−x^3 (x^3 +x)+x(x^3 +x)−x^2 +3x^3 =x^5 −x^3 +x−x^2 +3x^3 =x^2 (x^3 +x)−x^2 +x^3 +x =x^2 −x^2 +1 =1

$${x}^{\mathrm{8}} +\mathrm{3}{x}^{\mathrm{3}} \\ $$$$={x}^{\mathrm{8}} +{x}^{\mathrm{6}} −{x}^{\mathrm{6}} −{x}^{\mathrm{4}} +{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} \\ $$$$={x}^{\mathrm{5}} \left({x}^{\mathrm{3}} +{x}\right)−{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} +{x}\right)+{x}\left({x}^{\mathrm{3}} +{x}\right)−{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} \\ $$$$={x}^{\mathrm{5}} −{x}^{\mathrm{3}} +{x}−{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} \\ $$$$={x}^{\mathrm{2}} \left({x}^{\mathrm{3}} +{x}\right)−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x} \\ $$$$={x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$=\mathrm{1} \\ $$

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