\(\displaystyle x^{3} + 3x^{2} y + 3 xy^{2} + y^{3}\) is recognizable as the cube of the binomial \(\displaystyle x+y\). That is,

\(\displaystyle x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} = (x+y) ^{3}\)

Therefore, setting \(\displaystyle x = 6+ 3i\) and \(\displaystyle y= 6 - 3i\) and evaluating:

\(\displaystyle x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} =[ (6+3i) + (6-3i) ]^{3}\)

\(\displaystyle = (6+6+3i -3i) ^{3}\)

\(\displaystyle =12 ^{3}\)

\(\displaystyle = 1,728\).

\(\displaystyle x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}\) is recognizable as the cube of the binomial \(\displaystyle x-y\). That is,

\(\displaystyle x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}= (x-y) ^{3}\)

Therefore, setting \(\displaystyle x = 7+ 4i\) and \(\displaystyle y= 7 - 4i\) and evaluating:

\(\displaystyle x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}=[ (7+4i) - (7-4 i) ]^{3}\)

\(\displaystyle = (7-7 +4i+4i ) ^{3}\)

\(\displaystyle =(8i)^{3}\)

Applying the Power of a Product Rule and the fact that \(\displaystyle i^{3} = -i\):

\(\displaystyle (8i)^{3} = 8 ^{3} i ^{3} = 512 (-i) = -512i\),

the correct value.

To raise any expression \(\displaystyle A-B\) to the third power, use the pattern

\(\displaystyle \left (A+B \right )^{3} = A^{3} - 3A^{2} B + 3 AB^{2} - B^{3}\)

Setting \(\displaystyle A = 6, B = i \sqrt{7}\):

\(\displaystyle (6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot ( i \sqrt{7})^{2} - ( i \sqrt{7})^{3}\)

Taking advantage of the Power of a Product Rule:

\(\displaystyle (6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot i^{2} \cdot ( \sqrt{7})^{2} - i ^{3}\cdot (\sqrt{7})^{3}\)

Since \(\displaystyle i^{2} = -1\),

and

\(\displaystyle i^{3} = i^{2} \cdot i = -1 \cdot i = -i\):

\(\displaystyle (6- i \sqrt{7} )^{3}=216 - 3\cdot 36 \cdot i \sqrt{7} + 3 \cdot 6 \cdot 7 \cdot (-1) - (-i) \cdot 7\cdot \sqrt{7}\)

\(\displaystyle (6- i \sqrt{7} )^{3} =216 - 108i \sqrt{7} -126 + 7i\sqrt{7}\)

Collecting real and imaginary terms:

\(\displaystyle (6- i \sqrt{7} )^{3} =216 -126 - 108i \sqrt{7} + 7i\sqrt{7}\)

\(\displaystyle (6- i \sqrt{7} )^{3} =90- 101i \sqrt{7}\)

To raise any expression \(\displaystyle A-B\) to the third power, use the pattern

\(\displaystyle \left (A+B \right )^{3} = A^{3} - 3A^{2} B + 3 AB^{2} - B^{3}\)

Setting \(\displaystyle A = 6, B = 7i\):

\(\displaystyle (6-7i)^{3} = 6^{3} - 3\cdot 6^{2} \cdot 7i + 3 \cdot 6 \cdot (7i)^{2} - (7i)^{3}\)

Taking advantage of the Power of a Product Rule:

\(\displaystyle (6-7i)^{3} = 6^{3} - 3\cdot 6^{2} \cdot 7i + 3 \cdot 6 \cdot 7^{2} \cdot i^{2} - 7^{3} \cdot i^{3}\)

Since \(\displaystyle i^{2} = -1\),

and

\(\displaystyle i^{3} = i^{2} \cdot i = -1 \cdot i = -i\):

\(\displaystyle (6-7i)^{3} =216 - 3\cdot 36 \cdot 7i + 3 \cdot 6 \cdot 49 \cdot (-1) - 343 \cdot (-i)\)

\(\displaystyle (6-7i)^{3} =216 - 756 i -882 + 343 i\)

Collecting real and imaginary terms:

\(\displaystyle (6-7i)^{3} =216 -882 - 756 i + 343 i\)

\(\displaystyle (6-7i)^{3} =-666 - 413 i\)

Apply the Power of a Product Rule:

\(\displaystyle \left ( 4i\right )^{3} \cdot (3i)^{4}\)

\(\displaystyle = 4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}\)

\(\displaystyle i^{3} = i^{2} \cdot i = (-1 ) \cdot i = -i\),

and

\(\displaystyle i^{4} = (i^{2})^{2} = (-1)^{2} = 1\), 

so, substituting and evaluating:

\(\displaystyle 4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}\)

\(\displaystyle = 64 \cdot (-i) \cdot 81 \cdot 1\)

\(\displaystyle = - 64 \cdot 81 \cdot 1 \cdot i\)

\(\displaystyle =-5,184i\)

The easiest way to find \(\displaystyle \left (8 + i \sqrt{5} \right )^{4}\) is to note that  

\(\displaystyle \left (8 + i \sqrt{5} \right )^{4} =\left [ \left (8 + i \sqrt{5} \right )^{2} \right ]^{2}\).

Therefore, we can find the fourth power of \(\displaystyle 8 + i \sqrt{5}\) by squaring \(\displaystyle 8 + i \sqrt{5}\), then squaring the result.

Using the binomial square pattern to square \(\displaystyle 8 + i \sqrt{5}\):

\(\displaystyle (8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (i \sqrt{5} ) ^{2}\)

Applying the Power of a Product Property:

\(\displaystyle (8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + i^{2} \cdot \left ( \sqrt{5} \right) ^{2}\)

Since \(\displaystyle i^{2} = -1\) by definition: 

\(\displaystyle (8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (-1) \cdot 5\)

\(\displaystyle (8 + i \sqrt{5} )^{2} = 64 +16 i \sqrt{5} -5\)

\(\displaystyle = 59 +16 i \sqrt{5}\)

Square this using the same steps:

\(\displaystyle (59 +16 i \sqrt{5} )^{2} = 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + (16i \sqrt{5} ) ^{2}\)

\(\displaystyle = 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + 16^{2} \cdot i^{2} \cdot (\sqrt{5} ) ^{2}\)

\(\displaystyle =3,481+1,888 i \sqrt{5} + 256 \cdot (-1) \cdot 5\)

\(\displaystyle =3,481+1,888 i \sqrt{5} -1,280\)

\(\displaystyle =2,201+1,888 i \sqrt{5}\),

the correct response.

Apply the Power of a Product Rule:

\(\displaystyle \left ( 2 i \sqrt{5}\right )^{5} \cdot \left ( 3 i \sqrt{5} \right )^{3}\)

\(\displaystyle = 2 ^{5} \cdot i^{5} \cdot \left ( \sqrt{5}\right )^{5} \cdot 3 ^{3} \cdot i ^{3} \cdot \left ( \sqrt{5} \right )^{3}\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}\)

Applying the Product of Powers Rule:

\(\displaystyle 2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3} \cdot i^{5+3} \cdot \left ( \sqrt{5}\right )^{5+3}\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}\)

\(\displaystyle i\) raised to any multiple of 4 is equal to 1, and \(\displaystyle \sqrt{5} = 5 ^{\frac{1}{2}}\), so, substituting and evaluating:

\(\displaystyle 2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3}\cdot \left ( 5^{\frac{1}{2}} \right )^{8} \cdot 1\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3}\cdot 5^{ \frac{1}{2} \cdot 8} \cdot 1\)

\(\displaystyle = 2 ^{5} \cdot 3 ^{3}\cdot 5^{4} \cdot 1\)

\(\displaystyle = 32 \cdot 27 \cdot 625 \cdot 1\)

\(\displaystyle =540,000\)

This is not among the given choices.

\(\displaystyle x^{2} + 2xy + y^{2}\) conforms to the perfect square trinomial pattern

\(\displaystyle x^{2} + 2xy + y^{2} = (x+y) ^{2}\).

The easiest way to solve this problem is to add \(\displaystyle x\) and \(\displaystyle y\), then square the sum. 

The complex conjugate of a complex number \(\displaystyle a+bi\) is \(\displaystyle a-bi\).

\(\displaystyle x = 7 + 2 i \sqrt{5}\),

so \(\displaystyle y\) is the complex conjugate of this; 

\(\displaystyle y = 7 - 2 i \sqrt{5}\), 

and 

\(\displaystyle x+y = ( 7 + 2 i \sqrt{5} ) + ( 7 - 2 i \sqrt{5}) = 7+ 7+ 2 i \sqrt{5} - 2 i \sqrt{5} = 14\)

Substitute 14 for \(\displaystyle x+y\):

\(\displaystyle x^{2} + 2xy + y^{2} = (x+y) ^{2} = 14^{2} = 196\).

\(\displaystyle x^{2} + 2xy + y^{2}\) conforms to the perfect square trinomial pattern

\(\displaystyle x^{2} + 2xy + y^{2} = (x+y) ^{2}\).

The easiest way to solve this problem is to add \(\displaystyle x\) and \(\displaystyle y\), then square the sum. 

The complex conjugate of a complex number \(\displaystyle a+bi\) is \(\displaystyle a-bi\).

\(\displaystyle x = 4+ 7i\),

so \(\displaystyle y\) is the complex conjugate of this; 

\(\displaystyle y = 4 - 7i\), 

and 

\(\displaystyle x+y = (4+7i) + (4-7i) = 4+4 +7i - 7i = 8\)

Substitute 8 for \(\displaystyle x+y\):

\(\displaystyle x^{2} + 2xy + y^{2} = (x+y) ^{2} = 8^{2} = 64\).

\(\displaystyle x^{2} - 2xy + y^{2}\) conforms to the perfect square trinomial pattern

\(\displaystyle x^{2} - 2xy + y^{2} = (x-y) ^{2}\).

The easiest way to solve this problem is to subtract \(\displaystyle x\) and \(\displaystyle y\), then square the difference. 

The complex conjugate of a complex number \(\displaystyle a+bi\) is \(\displaystyle a-bi\).

\(\displaystyle x = 5+ 8i\),

so \(\displaystyle y\) is the complex conjugate of this; 

\(\displaystyle y = 5 - 8i\)

\(\displaystyle x-y = (5+8i) - (5-8i) = 5- 5 + 8i + 8i = 16i\)

Substitute \(\displaystyle 16i\) for \(\displaystyle x-y\):

\(\displaystyle x^{2} - 2xy + y^{2} = (x-y) ^{2} =( 16 i)^{2} = 16 ^{2} \cdot i^{2}\)

By definition, \(\displaystyle i^{2} = -1\), so, substituting,

\(\displaystyle 16 ^{2} \cdot i^{2} = 256 (-1 ) = -256\),

the correct choice.