Find a degree 3 polynomial having zeros $-6,1$ and 7 and the coefficient of $x^{3}$ equal 1 . The polynomial is
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Step 1 ：We are given the zeros of the polynomial as $-6,1$ and $7$ and the coefficient of $x^{3}$ is 1.

Step 2 ：A polynomial of degree 3 with zeros at $-6,1$ and $7$ can be written in the form $f(x) = a(x - r)(x - s)(x - t)$ where $r,s,t$ are the roots of the polynomial and $a$ is the coefficient of $x^{3}$.

Step 3 ：Substituting the given values, we get $f(x) = 1(x - (-6))(x - 1)(x - 7)$.

Step 4 ：Simplifying the above expression, we get the polynomial $x^{3} - 2x^{2} - 41x + 42$.

Step 5 ：This polynomial has the required roots and the coefficient of $x^{3}$ is 1, which matches the conditions given in the question.

Step 6 ：The degree 3 polynomial having zeros at $-6,1$ and $7$ and the coefficient of $x^{3}$ equal to 1 is \(\boxed{x^{3} - 2x^{2} - 41x + 42}\).