If x^{2}+3x+1=0, then the value of x^{3}+frac | vedclass

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(A) Given equation: $x^{2}+3x+1=0$Dividing the entire equation by $x$ (assuming $x 
eq 0$):$x+3+\frac{1}{x}=0$Rearranging the terms,we get:$x+\frac{1

Vedclass · Accepted Answer

$-18$

Vedclass · Answer

(A) Given equation: $x^{2}+3x+1=0$Dividing the entire equation by $x$ (assuming $x 
eq 0$):$x+3+\frac{1}{x}=0$Rearranging the terms,we get:$x+\frac{1}{x}=-3$Now,cubing both sides of the equation:$\left(x+\frac{1}{x}ight)^{3}=(-3)^{3}$Using the algebraic identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$:$x^{3}+\frac{1}{x^{3}}+3\left(x\cdot\frac{1}{x}ight)\left(x+\frac{1}{x}ight)=-27$Substituting the value of $(x+\frac{1}{x})=-3$:$x^{3}+\frac{1}{x^{3}}+3(1)(-3)=-27$$x^{3}+\frac{1}{x^{3}}-9=-27$$x^{3}+\frac{1}{x^{3}}=-27+9$$x^{3}+\frac{1}{x^{3}}=-18$

If $x^{2}+3x+1=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is

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