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The real root of the equation $x^{3}-6 x+9=0$ is
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1902 Upvotes
Verified Answer
The correct answer is:
$-3$
We have,
$$
\begin{gathered}
x^{3}-6 x+9=0 \\
\Rightarrow \quad(x+3)\left(x^{2}-3 x+3\right)=0 \\
\Rightarrow \quad x+3=0 \quad\left[\because x^{2}-3 x+3 \neq 0, x \in R\right] \\
\Rightarrow \quad x=-3
\end{gathered}
$$
$$
\begin{gathered}
x^{3}-6 x+9=0 \\
\Rightarrow \quad(x+3)\left(x^{2}-3 x+3\right)=0 \\
\Rightarrow \quad x+3=0 \quad\left[\because x^{2}-3 x+3 \neq 0, x \in R\right] \\
\Rightarrow \quad x=-3
\end{gathered}
$$
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