Polynomials — Maths STD 9 — Question

Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
Now, $A + B + C = 0 ($given$)$
$\text { If } A+B+C=0 \text {, then } A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C \ldots \text { (1) }$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
$(a+b+c)^3=27 a b c$
Hence, correct option is $(b).$