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**Hint:**according to the question we have to find the value of $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$when $z = x + iy$ is a complex number such that ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$

So, first of all we have to taking cube both side of the given expression ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$and put the conjugate of $z$ that is $\overline z = x - iy$

Formula used for the cube of $\left( {a + b} \right)$ that is mentioned below.

**Formula used:**

${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b).............................(A)$

Now, we have to compare both real and imaginary roots of $x - iy$and the expression obtained after taking the cube of $a + ib$to get the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$

Now, we have to put the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$in the given expression $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$to get the desired value.

**Complete answer:**

Step 1: First of all we have to taking cube both side of the given expression ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$

$ \Rightarrow \overline z = {(a + ib)^3}$

Now, use the formula of cube (A) that is mentioned in the solution hint.

$

\Rightarrow (x - iy) = {a^3} + {\left( {ib} \right)^3} + 3\left( a \right)\left( {ib} \right)\left( {a + ib} \right) \\

\Rightarrow (x - iy) = {a^3} + {i^3}{b^3} + 3{a^2}\left( {ib} \right) + 3a{\left( {ib} \right)^2} \\

$

As we know that ${i^2} = - 1$and ${i^3} = - i$

$

\Rightarrow (x - iy) = {a^3} - i{b^3} + 3{a^2}\left( {ib} \right) - 3a{b^2} \\

\Rightarrow (x - iy) = {a^3} - 3a{b^2} + i\left( {3{a^2}b - {b^3}} \right).......................(1) \\

$

Step 3: Now, we have to compare the both real and imaginary values of the expression (1) as obtained in the solution step 2.

$ \Rightarrow x = \left( {{a^3} - 3a{b^2}} \right)$and, $y = - \left( {\left( {3{a^2}b - {b^3}} \right)} \right)$

$ \Rightarrow \dfrac{x}{a} = \left( {{a^2} - 3{b^2}} \right)$and, $\dfrac{y}{b} = \left( {\left( { - 3{a^2} + {b^2}} \right)} \right)...............................(2)$

Step 4: Now, we have to the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$from the expression (2) in the given expression$\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$

\[

\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( {\left( {{a^2} - 3{b^2}} \right) + \left( { - 3{a^2} + {b^2}} \right)} \right) \\

\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( {{a^2} - 3{a^2} + {b^2} - 3{b^2}} \right) \\

\]

Now, solving the expression as obtained just above,

\[

\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( { - 2{b^2} - 2{a^2}} \right) \\

\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left\{ { - 2\left( {{a^2} + {b^2}} \right)} \right\} \\

\]

On eliminating the terms which can be eliminated,

\[ \Rightarrow - 2\]

**Hence, we have obtained the value of $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right) = - 2$. Therefore option (B) is correct.**

**Note:**

It is necessary that we have to find the cube of the expression given in the question then we can compare the real and imaginary roots with the conjugate of z which is $\overline z = x - iy$ and where z is $ = x + iy$.

It is necessary that we have to find the value of $\dfrac{x}{a}$and $\dfrac{y}{b}$with the help of comparing the expression of terms x and y.

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