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

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

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

$$\mathcal{T}hesolutions/answers(complete$$$$process)shouldbepostedas‘‘{\mathrm{Answer}}^{″}$$$$notascomment\left(kindly\right).$$

Commented by Gamil last updated on 19/Nov/22

Answered by Frix last updated on 18/Nov/22

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

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

$$x^3+x-1=0$$$$\mathcal{H}owhaveyoudividedwith0sir?$$

Commented by Frix last updated on 19/Nov/22

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

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

$$\mathcal{T}han\mathcal{X}\mathit{s}\mathit{i}\mathit{r}!$$

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

$$x^3+x=1;x^8+3x^3=?$$$$x^3=1-x$$$$▸x^8+3x^3={\left(x^3\right)}^2\cdot x^2+3x^3$$$$=x^2{(1-x)}^2+3(1-x)$$$$=x^2(1-2x+x^2)+3-3x$$$$=x^2-2x^3+x^3\cdot x+3-3x$$$$=x^2-2(1-x)+x(1-x)+3-3x$$$$=\cancel{x^2}-2+2x+x-\cancel{x^2}+3-3x$$$$=1$$$$$$

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

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

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