Question

If $x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$, then:

• A. $x^3+y^3+z^3=0$
• B. $x+y+z=27xyz$
• C. $(x+y+z{)}^3=27xyz$
• D. $x^3+y^3+z^3=27xyz$

Answer 1

The correct option is C $(x+y+z{)}^3=27xyz$

Recall the formula,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca\left)\right)$.

Lets substitute the values of $x,y,z$ as follows,

$x=a^3,y=b^3,z=c^3$.

So now,

$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=a+b+c=0$.

Using the formula above we can say that,

if $a+b+c=0\Rightarrow a^3+b^3+c^3=3abc$.

Substituting back, we get $x+y+z=3x^{1/3}{y}^{1/3}{z}^{1/3}$.

Cubing both sides we get, $(x+y+z{)}^3=27xyz$.

Therefore, option $C$ is correct.