# Awkward roots of a polynomial

Let there be a polynomial function, with integer coefficients, of least degree such that $7^\frac{1}3 + 49^\frac13$ is a root of the polynomial. What is the product of the roots of the $f(x) = 0$?

In problems which state the degree of the polynomial as least degree (with additional relevant information), is it assumed that we have to take the polynomial as degree 3 or 2? Considering how tedious the cubic polynomial solution is, I don't think that is to be used almost anywhere. How is one supposed to go about such a problem?

• Are you aware of Vieta's formulas? – Cataline Nov 22 '15 at 12:16
• Yes I am @Cataline – Sat D Nov 22 '15 at 12:17
• Try cubing the expression, and see what the result is. The minimum polynomial becomes clear after you have cubed it, i you are observant enough. – Cataline Nov 22 '15 at 12:20
• Ah damn, should have seen that. Thanks! – Sat D Nov 22 '15 at 12:22
• The least degree polynomial is $x-7^\frac{1}3 -49^\frac13$. – Bernard Nov 22 '15 at 12:37

Let $\alpha=7^{\frac13}$ so $\alpha^3=7$. We need to calculate the minimal polynomial of $x=\alpha+\alpha^2$. One has $x^3=\alpha^3+3\alpha^2\alpha^2+3\alpha\alpha^4+\alpha^6=7+3\cdot7\alpha+3\cdot7\alpha^2+49=56+21x$
Thus the searched minimal polynomial is $p(x)=x^3-21x-56$