$\displaystyle x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of 

$\displaystyle 1,\ -1,\ i,\ -i$ 

The product of $\displaystyle \left ( 5+2i \right )$ with each of these gives us:

$\displaystyle (5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$\displaystyle (5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$\displaystyle (5+2i)\cdot 1 = 5+2i$

$\displaystyle (5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

$\displaystyle [(-2+5i) + (2-5i)] + [(5+2i) + (-5-2i)] =$

$\displaystyle =[(-2)+2 +5i - 5i] + [5-5 + 2i - 2i] = [0] + [0] = 0$

What we notice is that each of the roots has a negative.  It thus makes sense that they will all cancel out.  Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$\displaystyle 1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$\displaystyle = (5+2i)\cdot[0] = 0$

Apply the Power of a Product Property:

$\displaystyle (3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of $\displaystyle i$ can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so 

$\displaystyle i^{6} = i^{2} = -1$

Substituting, 

$\displaystyle (3i)^{6} = 729 \cdot (-1) = -729$.

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:

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

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