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

\(\displaystyle a^{2}+b^{2}\).

Therefore, the product of \(\displaystyle 3 + i \sqrt {3 }\) and its complex conjugate \(\displaystyle 3 - i \sqrt {3 }\) can be found by setting \(\displaystyle a = 3\) and \(\displaystyle b = \sqrt{3}\) in this pattern:

\(\displaystyle a^{2}+b^{2} = 3^{2}+ (\sqrt{3})^{2} = 9 + 3 = 12\),

the correct response.

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

\(\displaystyle a^{2}+b^{2}\).

Therefore, the product of \(\displaystyle 11 - i \sqrt{11}\) and its complex conjugate \(\displaystyle 11 + i \sqrt{11}\) can be found by setting \(\displaystyle a = 11\) and \(\displaystyle b = \sqrt{11}\) in this pattern:

\(\displaystyle a^{2}+b^{2} = 11^{2}+ (\sqrt{11})^{2} = 121+11 = 132\),

the correct response.

The complex conjugate of a complex number \(\displaystyle a+bi\) is \(\displaystyle a - bi\), so \(\displaystyle \frac{3}{2} + \frac{1}{4} i\) has \(\displaystyle \frac{3}{2} - \frac{1}{4} i\) as its complex conjugate. 

The product of \(\displaystyle a+bi\) and \(\displaystyle a - bi\) is equal to \(\displaystyle a^{2}+ b^{2}\), so set \(\displaystyle a = \frac{3}{2} , b= \frac{1}{4}\) in this expression, and evaluate:

\(\displaystyle a^{2}+ b^{2} = \left ( \frac{3}{2} \right )^{2} + \left (\frac{1}{4} \right )^{2} = \frac{9}{4} + \frac{1}{16} = \frac{37}{16}\).

This is not among the given responses.

The two factors are both square roots of negative numbers, and are therefore imaginary. Write both in terms of \(\displaystyle i\) before multiplying:

\(\displaystyle \sqrt{-35} = i \sqrt{35}\)

\(\displaystyle \sqrt{-5} = i \sqrt{5}\)

Therefore, using the Product of Radicals rule:

 \(\displaystyle \sqrt{-35} \cdot \sqrt{-5}\)

\(\displaystyle = i \sqrt{35} \cdot i \sqrt{5}\)

\(\displaystyle = i \cdot i \cdot \sqrt{35} \cdot \sqrt{5}\)

\(\displaystyle = i ^{2}\cdot \sqrt{35 \cdot 5 }\)

\(\displaystyle =-1 \cdot \sqrt{175 }\)

\(\displaystyle =-1 \cdot \sqrt{25 }\cdot \sqrt{7}\)

\(\displaystyle =-1 \cdot 5\cdot \sqrt{7}\)

\(\displaystyle =-5\sqrt{7}\)

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