Question

If $x^2+4x+3=0$ then the value of $\frac{x^3}{x^6+27x^3+27}$ is

• A. $-1$
• B. $-\frac{1}{2}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-1$

Given that

$x^2+4x+3=0$

$\Rightarrow x^2+4x=-3$ .$\left(1\right)$

Cubing on both sides

$\Rightarrow (x^2+4x{)}^3=(-3{)}^3$

$\Rightarrow x^6+64x^3+12x^3(x^2+4x)=-27$

$\Rightarrow x^6+64x^3+12x^3(-3)=-27$ (from $\left(1\right)$)

$\Rightarrow x^6+64x^3-36x^3=-27$

$\Rightarrow x^6+28x^3=-27$

$\Rightarrow x^6+27x^3=-27-x^3$

$\Rightarrow x^6+27x^3+27=-x^3$

$\Rightarrow \frac{x^3}{x^6+27x^3+27}=-1$