Question

The cube root of $x+y+3x^{\frac{1}{3}}{y}^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})$ is

• A. $x+y$
• B. $x^{\frac{1}{3}}+y^{\frac{1}{3}}$
• C. $(x+y{)}^{\frac{1}{3}}$
• D. $(x+y{)}^3$

Answer 1

The correct option is B $x^{\frac{1}{3}}+y^{\frac{1}{3}}$

The given expression $x+y+3x^{\frac{1}{3}}{y}^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})$ can be rewritten as:

$x+y+3x^{\frac{1}{3}}{y}^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})={\left(x^{\frac{1}{3}}\right)}^3+{\left(y^{\frac{1}{3}}\right)}^3+3x^{\frac{1}{3}}{y}^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})={(x^{\frac{1}{3}}+y^{\frac{1}{3}})}^3(\because (a+b{)}^3=a^3+b^3+3ab(a+b))$

Now, the cube root can be determined as:

$\sqrt[3]{3{(x^{\frac{1}{3}}+y^{\frac{1}{3}})}^3}=x^{\frac{1}{3}}+y^{\frac{1}{3}}$

Hence, the cube root of $x+y+3x^{\frac{1}{3}}{y}^{\frac{1}{3}}(x^{\frac{1}{3}}+y^{\frac{1}{3}})$ is $x^{\frac{1}{3}}+y^{\frac{1}{3}}$.