Let \(\displaystyle x\) be a sixth root of 64. The question is to find a solution of the equation

\(\displaystyle x^{6} = 64\).

Subtracting 64 from both sides, this equation becomes

\(\displaystyle x^{6} - 64= 0\)

64 is a perfect square (of 8) The binomial at left can be factored first as the difference of two squares:

\(\displaystyle (x^{3} ) ^{2}- 8 ^{2}= 0\)

\(\displaystyle ( x^{3} + 8 )( x^{3} - 8 )= 0\)

8 is a perfect cube (of 2), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

\(\displaystyle x^{3} + 8 = x^{3}+ 2 ^{3} = (x+2) (x^{2}- x \cdot 2 + 2 ^{2}) = (x+2) (x^{2}-2x+4)\)

\(\displaystyle x^{3} - 8 = x^{3}+ 2 ^{3} = (x-2) (x^{2}+x \cdot 2 + 2 ^{2}) = (x-2) (x^{2}+2x+4)\)

The equation therefore becomes 

\(\displaystyle (x+2) (x^{2}-2x+4) (x-2) (x^{2}+2x+4) = 0\).

By the Zero Product Principle, one of these factors must be equal to 0.

If \(\displaystyle x+ 2 = 0\), then \(\displaystyle x = -2\); if \(\displaystyle x-2 = 0\), then \(\displaystyle x = 2\). Therefore, \(\displaystyle -2\) and 2 are sixth roots of 64. However, these are not choices, so we examine the other polynomials for their zeroes.

If \(\displaystyle x^{2}-2x+4 = 0\), then, setting \(\displaystyle a= 1, b= -2, c= 4\) in the following quadratic formula:

\(\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

\(\displaystyle = \frac{- (-2) \pm \sqrt{ (-2)^{2}-4(1)(4)}}{2(1)}\)

\(\displaystyle = \frac{2 \pm \sqrt{ 4 - 16 }}{2}\)

\(\displaystyle = \frac{2 \pm \sqrt{ -12 }}{2}\)

\(\displaystyle = \frac{2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}\)

\(\displaystyle = \frac{2 \pm 2i \sqrt{3 }}{2}\)

\(\displaystyle =1 \pm i \sqrt{3 }\)

If \(\displaystyle x^{2}+2x+4= 0\), then, setting \(\displaystyle a= 1, b=2, c= 4\) in the quadratic formula:

\(\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

\(\displaystyle = \frac{- 2 \pm \sqrt{ 2^{2}-4(1)(4)}}{2(1)}\)

\(\displaystyle = \frac{-2 \pm \sqrt{ 4 - 16 }}{2}\)

\(\displaystyle = \frac{-2 \pm \sqrt{ -12 }}{2}\)

\(\displaystyle = \frac{-2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}\)

\(\displaystyle = \frac{-2 \pm 2i \sqrt{3 }}{2}\)

\(\displaystyle =-1 \pm i \sqrt{3 }\)

Therefore, the set of sixth roots of 64 is 

\(\displaystyle \left \{ -4, 4,-1 - i \sqrt{3 }, -1 + i \sqrt{3 } ,1 - i \sqrt{3 }, 1 + i \sqrt{3 } \right \}\).

All four choices appear in this set.

Knowing the actual values of \(\displaystyle x\) and \(\displaystyle y\)  is not necessary to solve this problem. The product of the complex conjugates of two numbers is equal to the complex conjugate of the product of the numbers; that is,

 \(\displaystyle \overline{x} \cdot \overline{y} = \overline{xy}\)

\(\displaystyle xy = 13+ 7i\), so \(\displaystyle \overline{xy }= 13- 7i\), and

\(\displaystyle \overline{x} \cdot \overline{y} =13- 7i\),

which is not among the choices.

Applying the Power of a Product Rule:

\(\displaystyle x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3}\)

The complex conjugate of an imaginary number \(\displaystyle x = a+ bi\) is \(\displaystyle \overline{x} = a - bi\); the product of the two is 

\(\displaystyle x \overline{x} = a^{2} + b^{2}\)

\(\displaystyle x = 5 + 2i\), so, setting \(\displaystyle a = 5 , b = 2\) in the above pattern:

\(\displaystyle x \overline{x} = 5^{2} + 2^{2} = 25 + 4 = 29\)

Consequently, 

\(\displaystyle x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3} = 29 ^{3}= 24, 389\)

Follow the correct order of operations: parenthenses, exponents, multiplication, division, addition, subtraction.

\(\displaystyle \small (3+4)^2+(\frac{3+5}{2})+6\div 2\)

First, evaluate any terms in parenthesis.

\(\displaystyle (7)^2+(\frac{8}{2})+6\div 2\)

\(\displaystyle 7^2+4+6\div 2\)

Next, evaluate the exponent.

\(\displaystyle \small 49+4+6\div2\)

Divide.

\(\displaystyle \small 49+4+3\)

Finally, add.

\(\displaystyle \small 49+4+3=56\)

Follow the correct order of operations: parentheses, exponents, multiplication, division, addition, subtraction. (This is typically abbreviated as PEMDAS. Note that both multiplication and division, and addition and subtraction, are equal to each other in terms of rank, so when both are present, solving the equation proceeds from left to right).

First, simplify anything in parentheses.

\(\displaystyle \left(\frac{6}{6}\right)+8^2-4*6+5\)

\(\displaystyle \small 1+8^2-4*6+5\)

Next, simplify any terms with exponents.

\(\displaystyle \small 1+64-4*6+5\)

Now, perform multiplication.

\(\displaystyle \small 1+64-24+5\)

Since all we are left with is addition and subtraction, we perform simplification from left to right.

\(\displaystyle \small \small 65-24+5 = 41+5=46\)

Thus, our answer is:

\(\displaystyle \small \small 46\)

In modulo 7 arithmetic, a number is congruent to the remainder of its division by 7. 

Therefore, since \(\displaystyle 5 + 4 + 6 + 2 = 17\) and \(\displaystyle 17 \div 7 = 2 \textrm{ R }3\),

\(\displaystyle 5 + 4 + 6 + 2 \equiv 3 \mod 7\),

and the correct response is 3.

To solve \(\displaystyle 100+1.01+0.01+0.00001\), make sure the digits are aligned with the correct placeholder.  It is also possible to add term by term.

\(\displaystyle 100+1.01= 101.01\)

\(\displaystyle 101.01+0.01= 101.02\)

\(\displaystyle 101.02+ 0.00001=101.02001\)

The correct answer is: \(\displaystyle 101.02001\)

Step 1: Recall PEMDAS...

Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

Step 2: Perform the evaluation in separate pieces...

\(\displaystyle (2+1)^3=3^3=27\)

\(\displaystyle (3*4)=12\)

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

\(\displaystyle (24\div 6)=4\)

Step 3: Replace the values and keep the signs..

\(\displaystyle 27-12+4-4\)

Step 4: Evaluate:

\(\displaystyle (27-12)+(4-4)=15+0=15\)

Add all the ones digits.

\(\displaystyle 3+4+8+2+4 = 21\)

Add the tens digits with the two as the carryover.

\(\displaystyle 1+1+8+1+5+(2) = 18\)

Combine this value with the ones digit of the first number.

The answer is:  \(\displaystyle 181\)

Add the ones digits.

\(\displaystyle 4+9+9 = 22\)

Add the tens digits with the tens digit of the previous number as carryover.

\(\displaystyle 3+8+7+(2) = 20\)

Repeat the process with the hundreds digits.

\(\displaystyle 1+1+8+(2) = 12\)

Combine this number with the ones digits of the previous calculations.

The answer is:  \(\displaystyle 1202\)