jmc

We recognize the number $\sqrt[3]{7}+\sqrt[3]{49}$ from the difference-of-cubes factorization $$7-1=\left(\sqrt[3]{7}-1\right)\left(1+\sqrt[3]{7}+\sqrt[3]{49}\right).$$Solving for $\sqrt[3]{7}+\sqrt[3]{49},$ we get $$\sqrt[3]{7}+\sqrt[3]{49}=\frac{7-1}{\sqrt[3]{7}-1}-1=\frac{6}{\sqrt[3]{7}-1}-1.$$We can use this expression to build a polynomial which has $\sqrt[3]{7}+\sqrt[3]{49}$ as a root. First, note that $\sqrt[3]{7}$ is a root of $x^3-7=0.$ Then, $\sqrt[3]{7}-1$ is a root of $(x+1{)}^3-7=0,$ because $(\sqrt[3]{7}-1+1{)}^3-7=(\sqrt[3]{7}{)}^3-7=0.$ (You could also note that the graph of $y=(x+1{)}^3-7$ is a one-unit leftward shift of the graph of $y=x^3-7,$ so the roots of $(x+1{)}^3-7=0$ are one less than the roots of $x^3-7=0.$)

It follows that $\frac{6}{\sqrt[3]{7}-1}$ is a root of the equation $${\left(\frac{6}{x}+1\right)}^3-7=0,$$because when $x=\frac{6}{\sqrt[3]{7}-1},$ we have $\frac{6}{x}=\sqrt[3]{7}-1.$ We multiply both sides by $x^3$ to create the polynomial equation $$(6+x{)}^3-7x^3=0.$$Finally, replacing $x$ with $x+1$ like before, we see that $\frac{6}{\sqrt[3]{7}-1}-1$ is a root of the equation $$(7+x{)}^3-7(x+1{)}^3=0.$$This equation is equivalent to $$x^3-21x-56=0,$$so by Vieta's formulas, the product of the roots is $\boxed{56}.$