Let $a$ and $b$ be two distinct roots of the equation $x^3+3x^2-1=0$. The equation on which has $(ab)$ as its root is equal to

Let the three roots of the cubic equation x cubed plus 3x squared minus 1 equals 0 be a, b, and c. Using Vieta's formulas, the sum of the roots (a plus b plus c) equals negative 3, the sum of their products taken two at a time (ab plus bc plus ca) equals 0, and their product (abc) equals 1. We want a new equation whose roots are the pairwise products ab, bc, and ca. The sum of the new roots is ab plus bc plus ca equals 0, the product of the new roots taken two at a time is abc(a plus b plus c) equals 1 times negative 3 equals negative 3, and the product of the new roots is (abc) squared equals 1. Forming the cubic equation with these Vieta values gives x cubed minus 0x squared minus 3x minus 1 equals 0, which simplifies to x cubed minus 3x minus 1 equals 0.

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