The real root of the equation x^{3}-6x+9=0 is | vedclass

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(D) Given the equation: $x^{3}-6x+9=0$By testing integer factors of $9$,we find that for $x=-3$:$(-3)^{3}-6(-3)+9 = -27+18+9 = 0$Thus,$(x+3)$ is a fac

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$-3$

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(D) Given the equation: $x^{3}-6x+9=0$By testing integer factors of $9$,we find that for $x=-3$:$(-3)^{3}-6(-3)+9 = -27+18+9 = 0$Thus,$(x+3)$ is a factor of the polynomial.Dividing $x^{3}-6x+9$ by $(x+3)$,we get:$x^{3}+3x^{2}-3x^{2}-9x+3x+9 = x^{2}(x+3)-3x(x+3)+3(x+3) = (x+3)(x^{2}-3x+3) = 0$For the quadratic part $x^{2}-3x+3=0$,the discriminant $D = b^{2}-4ac = (-3)^{2}-4(1)(3) = 9-12 = -3$.Since $D Therefore,the only real root is $x=-3$.

The real root of the equation $x^{3}-6x+9=0$ is

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