Question
If $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}= 0,$ then:
✓
Answer
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$)$
If $A + B + C = 0,$ then $A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3ABC ...(1)$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation$,$ we get
$(a + b + c)^3 = 27abc$
Now$, A + B + C = 0 ($given$)$
If $A + B + C = 0,$ then $A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3ABC ...(1)$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation$,$ we get
$(a + b + c)^3 = 27abc$
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