We can set \(\displaystyle A = 2, B = 3i\) in the cube of a binomial pattern:

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

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

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

\(\displaystyle = 8 + 36i + 54 i^{2} + 27 i^{3}\)

\(\displaystyle = 8 + 36i + 54 (-1) + 27 (-i)\)

\(\displaystyle = 8 + 36i - 54 - 27 i\)

\(\displaystyle = -46 + 9i\)

Multiply these complex numbers out in the typical way:

\(\displaystyle {}(5+3i)(-2+i) = -10+5i-6i+3i^2\)

and recall that \(\displaystyle i^2=-1\) by definition. Then, grouping like terms we get

\(\displaystyle {} (-10-3)+(5i-6i) = -13-i\)

which is our final answer.

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

\(\displaystyle (5+8i)(6-2i)=30-10i+48i-16i^2\)

Remember that \(\displaystyle i=\sqrt-1\), so \(\displaystyle i^2=-1\).

Substitute in \(\displaystyle -1\) for \(\displaystyle i^2\)

\(\displaystyle 30-10i+48i-16i^2=30+38i-16(-1)=30+38i+16=46+38i\)

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

\(\displaystyle (2+5i)(-8-i)=-16-2i-40i-5i^2\)

Remember that \(\displaystyle i=\sqrt-1\), so \(\displaystyle i^2=-1\).

Substitute in \(\displaystyle -1\) for \(\displaystyle i^2\)

\(\displaystyle -16-2i-40i-5i^2=-16-42i-5(-1)=-16+5-42i=-11-42i\)

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

\(\displaystyle (2+6i)(5+3i)=10+6i+30i+18i^2\)

Remember that \(\displaystyle i=\sqrt-1\), so \(\displaystyle i^2=-1\).

Substitute in \(\displaystyle -1\) for \(\displaystyle i^2\)

\(\displaystyle 10+6i+30i+18i^2=10+36i+18(-1)=10-18+36i=-8+36i\)

Remember that 

\(\displaystyle i = \sqrt{-1}\\ i^2 = \sqrt{-1}^2 = -1\\ i^3 = i^2\cdot i = -1 \cdot i = -i\\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1\)

So the powers of \(\displaystyle i\) are cyclic. This means that when we try to figure out the value of an exponent of \(\displaystyle i\), we can ignore all the powers that are multiples of \(\displaystyle 4\) because they end up multiplying the end result by \(\displaystyle 1\), and therefore do nothing.

This means that 

\(\displaystyle ai^{93} + bi^{35} + ai^{24} - bi^{86}\\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2\)

Now, remembering the relationships of the exponents of \(\displaystyle i\), we can simplify this to:

\(\displaystyle ai - bi + a - (-b) = ai-bi+a+b\\ = (a+b) + (a-b)i = 40+20i\)

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships: 

\(\displaystyle a+b=40\)

\(\displaystyle a-b=10\)

No matter how you solve it, you get the values \(\displaystyle a=30\), \(\displaystyle b=10\).

Step 1: Split the \(\displaystyle \sqrt{-243}\) into \(\displaystyle \sqrt{243}\cdot\sqrt{-1}\).

Step 2: Recall that \(\displaystyle \sqrt{-1}=i\), so let's replace it.

We now have: \(\displaystyle \sqrt{243}\cdot i\).

Step 3: Simplify \(\displaystyle \sqrt{243}\). To do this, we look at the number on the inside.

\(\displaystyle 243=3\cdot3\cdot3\cdot3\cdot3\).

Step 4: Take the factorization of \(\displaystyle 243\) and take out any pairs of numbers. For any pair of numbers that we find, we only take \(\displaystyle 1\) of the numbers out.

We have a pair of \(\displaystyle 3's\), so a \(\displaystyle 3\) is outside the radical.
We have another pair of \(\displaystyle 3's\), so one more three is put outside the radical.

We need to multiply everything that we bring outside: \(\displaystyle 3*3=9\)

Step 5: The \(\displaystyle i\) goes with the 9...\(\displaystyle 9i\)

Step 6: The last \(\displaystyle 3\) after taking out pairs gets put back into a square root and is written right after the \(\displaystyle i\)

It will look something like this: \(\displaystyle 9i\sqrt{3}\)

There are two ways to simplify this problem: 

Method 1: 

\(\displaystyle We\ know\ that\ i^2=-1,\ when\ we\ replace\ that\ we\ get:\)

\(\displaystyle 5\sqrt{-12}\)

\(\displaystyle =5i\sqrt{12}(take\ the\ i\ out)\)

\(\displaystyle =5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)\)

\(\displaystyle =10i{\sqrt{3}}\ reduce\)

Method 2: 

\(\displaystyle We\ know\ that\ the\ square\ root\ of\ anything\ squared\ is\ just\ itself\)

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

\(\displaystyle 5\sqrt{12i^{2}}=5i\sqrt{12}\)

\(\displaystyle =5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)\)

\(\displaystyle =10i{\sqrt{3}}\ reduce\)

\(\displaystyle First\ simplify\ each\ radical:\)

\(\displaystyle \sqrt{-49}-\sqrt{-16}=7i-4i\)

\(\displaystyle Next\ combine\ like\ terms:\)

\(\displaystyle 7i-4i=3i\)

\(\displaystyle When\ reducing\ the\ square\ root\ of\ a\ negative\ number\ -\ we\ must\ first\ \)

 \(\displaystyle i=\sqrt{-1}\)

\(\displaystyle \sqrt{-72}=\sqrt{-1}\sqrt{72}=i\sqrt{72}\)

\(\displaystyle Once\ we\ have\ taken\ out\ the\ i,\ then\ we\ can\ simply\ break\ down\) \(\displaystyle the\ number\ under\ the\ radical\ by\ finding\ the\ largest\ perfect\ square\ that\ will go\ into\ it.\)

\(\displaystyle 36\ will\ go\ into\ 72\ so\ we\ break\ this\ down\ to\ become:\)

\(\displaystyle i\sqrt{36}\sqrt{2}=6i\sqrt{2}\)