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(D) Let the roots of the equation $x^3-13x^2+15x+189=0$ be $\alpha, \alpha+2,$ and $\beta$.From the relation between roots and coefficients:$1) \alpha

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$-3, 7, 9$

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(D) Let the roots of the equation $x^3-13x^2+15x+189=0$ be $\alpha, \alpha+2,$ and $\beta$.From the relation between roots and coefficients:$1) \alpha + (\alpha+2) + \beta = 13 \implies 2\alpha + \beta = 11 \implies \beta = 11 - 2\alpha$$2) \alpha(\alpha+2) + (\alpha+2)\beta + \alpha\beta = 15$$3) \alpha(\alpha+2)\beta = -189$Substitute $\beta = 11 - 2\alpha$ into the third equation:$\alpha(\alpha+2)(11-2\alpha) = -189$$(\alpha^2+2\alpha)(11-2\alpha) = -189$$11\alpha^2 - 2\alpha^3 + 22\alpha - 4\alpha^2 = -189$$-2\alpha^3 + 7\alpha^2 + 22\alpha + 189 = 0$$2\alpha^3 - 7\alpha^2 - 22\alpha - 189 = 0$Testing values,if $\alpha = -3$:$2(-27) - 7(9) - 22(-3) - 189 = -54 - 63 + 66 - 189 
eq 0$If $\alpha = 7$:$2(343) - 7(49) - 22(7) - 189 = 686 - 343 - 154 - 189 = 0$.So,$\alpha = 7$ is a root. The roots are $7, 7+2=9,$ and $\beta = 11-2(7) = -3$.The roots are $-3, 7, 9$.

The difference between two roots of the equation $x^3-13x^2+15x+189=0$ is $2$. Then the roots of the equation are:

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