Question

Let $p\left(x\right)=0$ be a polynomial equation of least possible degree, with rational coefficients, having $7^{1/3}+{49}^{1/3}$ as one of its roots. Then the product of all the roots of $p\left(x\right)=0$ is

• A. $7$
• B. $49$
• C. $56$
• D. $63$

Answer 1

The correct option is C $56$

$\Rightarrow$ $x=\sqrt[3]{37}+\sqrt[3]{349}$ --- [ Given ]

$\Rightarrow$ $x^3=(\sqrt[3]{37}+\sqrt[3]{349}{)}^3$

$\Rightarrow$ $x^3=(\sqrt[3]{37}{)}^3+(\sqrt[3]{349}{)}^3+3\times \left(7{)}^{1/3}\right(7{)}^{2/3}(\sqrt[3]{37}+\sqrt[3]{349})$

$\Rightarrow$ $x^3=7+49+3\times 7(\sqrt[3]{37}+\sqrt[3]{349})$

$\Rightarrow$ $x^3=56+3\times \left(x\right)$

$\Rightarrow$ $x^3=56+3x$

$\Rightarrow$ $x^3-3x-56=0$

$\Rightarrow$ This gives a cubic polynomial in x and product of roots $=56$