Question

If $x=\sqrt[3]{3a+\sqrt{a^2-b^3}}+\sqrt[3]{3a-\sqrt{a^2-b^3}}$ then $x^3+3bx=$ ____________.

• A. $2a$
• B. $2b$
• C. $3a$
• D. $4a$

Answer 1

The correct option is A $2a$

Given,

$x=\sqrt[3]{3a+\sqrt{a^2-b^3}}+\sqrt[3]{3a-\sqrt{a^2-b^3}}$.......(1).

Now cubing both sides we get,

$x^3=a+\sqrt{a^2-b^3}+a-\sqrt{a^2-b^3}-3$$\sqrt[3]{3a+\sqrt{a^2-b^3}}\sqrt[3]{3a-\sqrt{a^2-b^3}}$$(\sqrt[3]{3a+\sqrt{a^2-b^3}}+\sqrt[3]{3a-\sqrt{a^2-b^3}})$

or, $x^3=2a-3bx$ [ Using (1)and $\left(\sqrt[3]{3a+\sqrt{a^2-b^3}}\right)\left(\sqrt[3]{3a-\sqrt{a^2-b^3}}\right)=\sqrt[3]{3a^2-(a^2-b^3)}=b$]

or, $x^3+3bx=2a$.