Use FOIL to multiply complex numbers as follows:

$\displaystyle (2+3i)(1-i)$

$\displaystyle 2-2i+3i-3i^2$

Since $\displaystyle i= \sqrt{-1}$, it follows that $\displaystyle i^2 = -1$, so then:

$\displaystyle 2-2i+3i-3 \cdot (-1)$

$\displaystyle 2-2i+3i+3$

Combining like terms gives:

$\displaystyle 5+i$

Use FOIL:

$\displaystyle (3+2i)(3-2i)=9-6i+6i-4i^2$

Combine like terms:

$\displaystyle =9-4i^2$

But since $\displaystyle i^2=-1$, we know

$\displaystyle 9-4i^2=9-4 \cdot (-1)=9+4=13$

$\displaystyle x^{2} - 2xy + y^{2}$ 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 $\displaystyle a-bi$.

$\displaystyle x = 5+ i \sqrt{7}$,

so $\displaystyle y$ is the complex conjugate of this; 

$\displaystyle x = 5- i \sqrt{7}$

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

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

$\displaystyle =( 2i \sqrt{7} )^{2}$

Taking advantage of the Power of a Product Rule and the fact that $\displaystyle i^{2} = -1$:

$\displaystyle = 2^{2} \cdot i ^{2}\cdot (\sqrt{7}) ^{2}$

$\displaystyle =4 \cdot (-1 )\cdot 7$

$\displaystyle = -28$

By the Power of a Power Rule, the fourth power of any number is equal to the square of the square of that number:

$\displaystyle x^{4} = x ^{2 \cdot 2} =( x^{2})^{2}$

Therefore, one way to raise $\displaystyle 7+2i$ to the fourth power is to square it, then to square the result.

Using the binomial square pattern to square $\displaystyle 7+2i$:

$\displaystyle (7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + (2i ) ^{2}$

Applying the Power of a Product Property:

$\displaystyle (7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + 2^{2} \cdot i ^{2}$

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

$\displaystyle (7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + 2^{2} \cdot (-1)$

$\displaystyle = 49 + 28i -4$

$\displaystyle = 45 + 28i$

Square this using the same steps:

$\displaystyle (45 + 28i )^{2} = 45 ^{2} + 2 \cdot 45 \cdot 28 i + (28 i) ^{2}$

$\displaystyle = 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} i ^{2}$

$\displaystyle = 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} (-1)$

$\displaystyle = 2,025 +2,520 i -784$

$\displaystyle = 1,241 +2,520 i$

The easiest way to find $\displaystyle (8 - 3i)^{4}$ is to note that  

 $\displaystyle (8 - 3i)^{4} =[ (8 - 3i)^{2} ]^{2}$.

Therefore, we can find the fourth power of $\displaystyle 8 - 3i$ by squaring $\displaystyle 8 - 3i$, then squaring the result.

Using the binomial square pattern to square $\displaystyle 8 - 3i$:

$\displaystyle (8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + (3i)^{2}$

Applying the Power of a Product Property:

$\displaystyle (8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + 3^{2} \cdot i^{2}$

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

$\displaystyle (8 - 3i)^{2} = 64 - 48 i + 9 (-1)$

$\displaystyle = 64 - 48 i -9$

$\displaystyle = 55 - 48 i$

Square this using the same steps:

$\displaystyle ( 55 - 48 i )^{2} = 55^{2} - 2 \cdot 55 \cdot 48 i + (48i)^{2}$

$\displaystyle = 3,025-5,280 i + 48^{2} i^{2}$

$\displaystyle = 3,025-5,280 i + 2,304(-1)$

$\displaystyle = 3,025-5,280 i - 2,304$

$\displaystyle = 721-5,280 i$

Therefore, $\displaystyle (8 - 3i)^{4} = 721-5,280 i$

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 $\displaystyle A = 5, B = 4i$:

$\displaystyle (5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot (4i)^{2} + (4i)^{3}$

Taking advantage of the Power of a Product Rule:

$\displaystyle (5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot 4^{2} \cdot i^{2} +4 ^{3} \cdot i ^{3}$

Since $\displaystyle i^{2} = -1$ and $\displaystyle i^{3} = i^{2} \cdot i = -1 \cdot i = -i$:

$\displaystyle (5+4i)^{3} = 125 + 3\cdot 25 \cdot 4 i + 3 \cdot 5 \cdot 16 \cdot (-1) +64 \cdot (- i )$

$\displaystyle (5+4i)^{3} = 125+ 300i -240 -64i$

Collecting real and imaginary terms:

$\displaystyle (5+4i)^{3} = 125 -240 + 300i -64i$

$\displaystyle (5+4i)^{3} = -115 + 236i$

$\displaystyle a^{-m}$ is defined to be equal to $\displaystyle \frac{1}{a^{m}}$ for any real or imaginary $\displaystyle a$ and for any real $\displaystyle m$; therefore,

$\displaystyle i^{-31} = \frac{1}{i^{31}}$

To evaluate a positive power of $\displaystyle i$, divide the power by 4 and note the remainder:

$\displaystyle 31 \div 4= 7 \textrm{ R }3$

Therefore, 

$\displaystyle i ^{31} = i ^{3} = -i$

Substituting,

$\displaystyle i^{-31} = \frac{1}{-i}$

Rationalizing the denominator by multiplying both numerator and denominator by $\displaystyle i$:

$\displaystyle i^{-31} = \frac{1 \cdot i}{-i \cdot i }= \frac{ i}{-i ^{2}}= \frac{ i}{-(-1)}= \frac{ i}{1}= i$

Since $\displaystyle 2^{x}= 2^{6}$

Hence $\displaystyle x=6$

$\displaystyle \sqrt[3]{x^{9}xy^{6}y^{2}z^{9}}=x^{3}y^{2}z^{3}\sqrt[3]{xy^{2}}$

From the equation one can see that

$\displaystyle \sqrt{x} = 5$

Hence $\displaystyle x$ must be equal to 25.