Question

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Answer 1

We are given the following equation;

$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}=0$

So, For solving the question, let's subtract "$\text{c}^{\frac{1}{3}}$" from both the sides, i.e., the L.H.S. and the R.H.S.

$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}-\text{c}^{\frac{1}{3}}=0-\text{c}^{\frac{1}{3}}$

Simplify;

$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}=-\text{c}^{\frac{1}{3}}$

Now, let's cube both the sides of the above equation;

$(\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}})^3=(-\text{c}^{\frac{1}{3}})^3$

Simplify;  :)

$(\text{a}^{\frac{1}{3}})^3+(\text{b}^{\frac{1}{3}})^3+3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}})=-c$

Now as we know that "$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}=-\text{c}^{\frac{1}{3}}$", so let's replace $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}$ to $-\text{c}^{\frac{1}{3}}$.

So, our equation looks like;

$(\text{a}^{\frac{1}{3}})^3+(\text{b}^{\frac{1}{3}})^3+3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})=-c$

Now lets add c on both the sides;

$(\text{a}^{\frac{1}{3}})^3+(\text{b}^{\frac{1}{3}})^3+3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})+c=-c+c$

Let's Simplify;

$\text{a}+\text{b}+\text{c}+3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})=0$

Now let's subtract "$3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})$" from both the sides;

We get;

$\text{a}+\text{b}+\text{c}+3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})-3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})=0-3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})$

Simplifying the above equation;

$\text{a}+\text{b}+\text{c}=-3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot(-\text{c}^{\frac{1}{3}})$

Let's get rid of the parenthesis from the R.H.S.;

$\text{a}+\text{b}+\text{c}=3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot\text{c}^{\frac{1}{3}}$

As we got a much simpler equation, let's cube both the sides;

$(\text{a}+\text{b}+\text{c})^3=(3\cdot\text{a}^{\frac{1}{3}}\cdot\text{b}^{\frac{1}{3}}\cdot\text{c}^{\frac{1}{3}})^3$

When we simplify the above equation, we get;

$(\text{a}+\text{b}+\text{c})^3=(3\text{a}\text{b}\text{c}^{\frac{1}{3}})^3$

And Simplifying it further, we get;

$\bold{(a+b+c)^3=27abc}$

Now let's check the options,

We got the option (b) as the answer.