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The number of real roots of the equation $\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0,(x \neq 0)$ is
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The correct answer is:
$0$
Given equation,
$\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0(x \neq 0)$
$\Rightarrow \quad \frac{3\left(x^2+1\right)^3+x^2\left(x^2+1\right)}{3 x^3}=0$
$\Rightarrow \quad\left(x^2+1\right)\left[3\left(x^2+1\right)^2+x^2\right]=0$ ...(i)
$\begin{aligned} & \because \quad x^2 \geq 0 \\ & \Rightarrow \quad\left(x^2+1\right) \geq 1 \text { and } 3\left(x^2+1\right)^2+x^2 \geq 3\end{aligned}$
By using this result in Eq. (i), we can observe that
for no real value of $x$ will satisfy Eq. (i).
Hence, the number of real roots is zero.
$\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0(x \neq 0)$
$\Rightarrow \quad \frac{3\left(x^2+1\right)^3+x^2\left(x^2+1\right)}{3 x^3}=0$
$\Rightarrow \quad\left(x^2+1\right)\left[3\left(x^2+1\right)^2+x^2\right]=0$ ...(i)
$\begin{aligned} & \because \quad x^2 \geq 0 \\ & \Rightarrow \quad\left(x^2+1\right) \geq 1 \text { and } 3\left(x^2+1\right)^2+x^2 \geq 3\end{aligned}$
By using this result in Eq. (i), we can observe that
for no real value of $x$ will satisfy Eq. (i).
Hence, the number of real roots is zero.
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