Multiple choice

Let $p(x)=0$ be a polynomial equation of the least possible degree, with rational coefficients, having $\sqrt [3]{7}+\sqrt [3]{49}$ as one of its roots, then the product of all the roots of $p(x)=0$, is:

  1. $56$
  2. $63$
  3. $7$
  4. $49$
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A Correct answer
Explanation

Let x = 7^(1/3) + 49^(1/3). Cubing both sides: x^3 = 7 + 49 + 3 * 7^(1/3) * 49^(1/3) * (7^(1/3) + 49^(1/3)) = 56 + 3 * 7 * x. So x^3 - 21x - 56 = 0. The product of the roots of a cubic ax^3 + bx^2 + cx + d = 0 is -d/a. Here, -(-56)/1 = 56.

AI explanation

Let the root be x = 7^(1/3) + 7^(2/3), and by cubing both sides we get x^3 = 7 + 49 + 3*(7*x). This simplifies to the minimal polynomial equation x^3 - 21x - 56 = 0. Based on this polynomial, the product of all its roots is given by the constant term divided by the leading coefficient with a sign change of (-1)^n. The product is calculated as -(-56)/1, which equals 56.

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