Use the properties of powers to simplify the expression:

\(\displaystyle \left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}\)

\(\displaystyle = \left (x^{2} \right )^{3} y ^{3} \cdot x ^{3}\left ( y ^{2} \right )^{3}\)

\(\displaystyle = \left (x^{2} \right )^{3} \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2} \right )^{3}\)

\(\displaystyle = \left (x^{2 \cdot 3} \right ) \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2\cdot 3} \right )\)

\(\displaystyle = x^{6} \cdot x ^{3}\cdot y ^{3} \cdot y ^{6}\)

\(\displaystyle = x^{6+3} \cdot y ^{3+ 6}\)

\(\displaystyle = x^{9} y ^{9} = (xy)^{9}\)

(a) If \(\displaystyle x = 6, y = 10\), then 

\(\displaystyle \left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 6 \cdot 10)^{9} = 60^{9}\)

(b) If \(\displaystyle x = y = 8\), then 

\(\displaystyle \left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 8 \cdot 8)^{9} = 64^{9}\)

(b) is greater.

Use the square of a binomial pattern as follows:

\(\displaystyle \left (100 + \sqrt{x} \right )^{2}\)

\(\displaystyle = 100^{2}+ 2 \cdot 100 \cdot \sqrt {x} +(\sqrt{x})^{2}\)

\(\displaystyle = 10,000+ 2 00 \sqrt {x} +x\)

This expression is not equivalent to any of the choices.

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

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

\(\displaystyle = x^{2}+ 6 x + 9\)

\(\displaystyle y = 16z\), so

\(\displaystyle 16z = x^{2}+ 6 x + 9\)

\(\displaystyle \frac{16z}{16} = \frac{x^{2}+ 6 x + 9}{16}\)

\(\displaystyle z = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}\)

\(\displaystyle z = w - 6\), so 

\(\displaystyle w = z+ 6\)

\(\displaystyle = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}+6\)

\(\displaystyle = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16} + \frac{96}{16}\)

\(\displaystyle = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{105}{16}\)

By the Power of a Power Principle, 

\(\displaystyle (a ^{3} )^{2} = a ^{3 \cdot 2 } = a ^{6}\)

Therefore, 

\(\displaystyle a ^{6} = (a ^{3} )^{2} = (-17)^{2} = 17^{2} = 17 \cdot 17 = 289\)

It follows that \(\displaystyle a ^{6} > - 34\)

By the Power of a Power Principle, 

\(\displaystyle (t ^{3} ) ^{2} = t^{3 \cdot 2 } = t ^{6} = 121\) 

Therefore, \(\displaystyle t ^{3}\) is a square root of 121, of which there are two - 11 and \(\displaystyle -11\). Since it is possible for a third (odd-numbered) power of a real number to be positive or negative, we cannot eliminate either possibility, so either

\(\displaystyle t = 11\)

or 

\(\displaystyle t = -11 < 11\).

Therefore, we cannot determine whether \(\displaystyle t\) is less than 11 or equal to 11.

By the Power of a Product Principle, 

\(\displaystyle (xy) ^{4} = x^{4} \cdot y ^{4}\)

Also, by the Power of a Power Principle

\(\displaystyle y^{4} = y^{2 \cdot 2} = (y^{2})^{2}\)

Therefore, 

\(\displaystyle (xy) ^{4} = x^{4} \cdot (y^{2})^{2} = 12 \cdot 8 ^{2} = 12 \cdot 64 = 768\)

Any nonzero number raised to an even power, such as 4, is a positive number. Therefore, 

\(\displaystyle -16 a ^{4} = -16 \cdot a ^{4}\) is the product of a negative number and a positive number, and is therefore negative. 

By the same reasoning,  \(\displaystyle (-2 a )^{4}\) is a positive number.

It follows that \(\displaystyle (-2 a )^{4} > -16 a ^{4}\).

By the Power of a Power Principle,

\(\displaystyle (t ^{3} )^{2} = t ^{3 \cdot 2 } = t ^{6}\)

By way of the Power of a Quotient Principle, 

\(\displaystyle \left ( \frac{2}{t} \right ) ^{6} = \frac{2^{6}}{t^{6} } = \frac{2^{6}}{(t^{3}) ^{2}} = \frac{64}{576} = \frac{64 \div 64 }{576 \div 64} = \frac{1}{9}\).

\(\displaystyle (2t - 3u) (2t+3u )\), as the product of a sum and a difference, can be rewritten using the difference of squares pattern:

\(\displaystyle (2t - 3u) (2t+3u )\)

\(\displaystyle =(2t )^{2}-( 3u)^{2}\)

\(\displaystyle =2^{2} \cdot t ^{2}- 3^{2}\cdot u^{2}\)

\(\displaystyle =2^{2}t ^{2}- 3^{2}u^{2}\)

\(\displaystyle =4t ^{2}- 9u^{2}\)

By the Power of a Power Principle, 

\(\displaystyle (t^{2} ) ^{2} = t^{2 \cdot 2 } = t^{4}\)

Therefore, \(\displaystyle t^{2}\) is a square root of \(\displaystyle t^{4}\) - that is, a square root of 121. 121 has two square roots, \(\displaystyle -121\) and 121, but since \(\displaystyle t\) is real, \(\displaystyle t ^{2}\) must be the positive choice, 11. Similarly, \(\displaystyle u^{2}\) is the positive square root of 81, which is 9.

The above expression can be evaluated as

\(\displaystyle 4t ^{2}- 9u^{2} = 4 (11) - 9 (9) = 44 - 81 = -37\).

Multiply the polynomials through distribution:

\(\displaystyle (t- u) (t^{2}+ tu+u^{2})\)

\(\displaystyle = t (t^{2}+ tu+u^{2}) - u (t^{2}+ tu+u^{2})\)

\(\displaystyle = t \cdot t^{2}+ t \cdot tu+ t \cdot u^{2} - u \cdot t^{2}- u \cdot tu - u \cdot u^{2}\)

\(\displaystyle = t^{3}+ t^{2} u+ t u^{2} - t^{2} u - tu^{2} - u^{3}\)

\(\displaystyle = t^{3}+ t^{2} u- t^{2} u + t u^{2} - tu^{2} - u^{3}\)

\(\displaystyle = t^{3} - u^{3}\)

The absolute value of \(\displaystyle t\) is 4, so either \(\displaystyle t = -4\) or \(\displaystyle t = 4\). Likewise, \(\displaystyle u = -3\) or \(\displaystyle u = 3\). 

If \(\displaystyle t = -4\) and \(\displaystyle u = 3\), we see that 

\(\displaystyle (t- u) (t^{2}+ tu+u^{2})\)

\(\displaystyle = t^{3} - u^{3}\)

\(\displaystyle =(-4 )^{3} - 3^{3}\)

\(\displaystyle = -64 - 27 = -91\)

If \(\displaystyle t = 4\) and \(\displaystyle u = -3\), we see that 

\(\displaystyle (t- u) (t^{2}+ tu+u^{2})\)

\(\displaystyle = t^{3} - u^{3}\)

\(\displaystyle =4^{3} - (- 3)^{3}\)

\(\displaystyle = 64 -( - 27 )= 91\)

In the first scenario, \(\displaystyle (t- u) (t^{2}+ tu+u^{2}) < 37\); in the second, \(\displaystyle (t- u) (t^{2}+ tu+u^{2}) > 37\). This makes the information insufficient.