To multiply a difference squared, square the first term and add two times the multiplication of the two terms. Then add the second term squared.

\(\displaystyle (3x)^2+2(3x)(5)+(5)^2=9x^2+30x+25\)

Use the square of a sum pattern, substituting \(\displaystyle 4x\) for \(\displaystyle A\) and \(\displaystyle 7y\) for \(\displaystyle B\) in the pattern:

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

\(\displaystyle (4x + 7y)^{2}= (4x)^{2}+ 2 (4x) (7y) + (7y)^{2}\)

\(\displaystyle = 16x^{2}+ 56xy + 49y^{2}\)

Use the square of a sum pattern, substituting \(\displaystyle 5 \sqrt{x}\) for \(\displaystyle A\) and \(\displaystyle 3 \sqrt{y}\) for \(\displaystyle B\) in the pattern:

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

\(\displaystyle (5 \sqrt{x}+3 \sqrt{y})^{2}= (5 \sqrt{x})^{2}+ 2 (5 \sqrt{x})\left (3 \sqrt{y} \right ) + \left (3 \sqrt{y} \right )^{2}\)

\(\displaystyle (5 \sqrt{x}+3 \sqrt{y})^{2}= 25x+ 30 \sqrt{xy} + 9y\)

or 

\(\displaystyle 25x + 9y+ 30 \sqrt{xy}\)

Multiply vertically as follows:

                    \(\displaystyle \begin{matrix} x^{2}+ \; \; \; x +\; \; \; 7\\\underline{x^{2}+\; \; \; x +\; \; \; 7} \end{matrix}\)

                    \(\displaystyle 7 x^{2}+7 x +49\)

          \(\displaystyle x^{3}+ x^{2} + 7x\)

\(\displaystyle \underline{x^{4}+ x^{3}+7 x^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }\)

\(\displaystyle x^{4}+2x^{3}+15x^{2}+14x+49\)

Use the square of a sum pattern, substituting \(\displaystyle y\) for \(\displaystyle A\) and \(\displaystyle \sqrt{17}\) for \(\displaystyle B\) in the pattern:

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

\(\displaystyle (y + \sqrt{17})^{2}= y^{2}+ 2y \sqrt{17} + (\sqrt{17})^{2}\)

\(\displaystyle (y + \sqrt{17})^{2}= y^{2}+ 2y \sqrt{17} + 17\)

This is not equivalent to any of the given choices.

Use the square of a sum pattern, substituting \(\displaystyle 5 \sqrt{x}\) for \(\displaystyle A\) and \(\displaystyle 3 \sqrt{y}\) for \(\displaystyle B\) in the pattern:

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

\(\displaystyle \left (\frac{1}{4}x + \frac{1}{7} \right )^{2}= \left (\frac{1}{4}x \right )^{2}+ 2 \left (\frac{1}{4}x \right )\left (\frac{1}{7} \right ) + \left (\frac{1}{7} \right )^{2}\)

\(\displaystyle = \frac{1}{16}x ^{2}+ \frac{1}{14}x +\frac{1}{49}\)

Use the square of a sum pattern, substituting \(\displaystyle 3.7x\) for \(\displaystyle A\) and \(\displaystyle 1.4 y\) for \(\displaystyle B\) in the pattern:

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

\(\displaystyle \left (3.7x + 1.4 y \right )^{2}= (3.7x)^{2}+ 2 (3.7x)(1.4 y) + (1.4 y) ^{2}\)

\(\displaystyle = 13.69x^{2}+ 10.36x y +1.96 y ^{2}\)

Simplify the base of 9 and 27 in order to have a common base.

(3^x)(9)=27²

= (3^x)(3²)=(3³)²

=(3^x+2)=3⁶

Therefore:

x+2=6

x=4

The terms of \(\displaystyle 4x^{4}+ 36 x^{2}\) have \(\displaystyle 4x^{2}\) as their greatest common factor, so

\(\displaystyle 4x^{4}+ 36 x^{2} = 4x^{2} (x^{2}+9)\)

\(\displaystyle x^{2}+ 9\) is a prime polynomial. 

Of the five choices, only \(\displaystyle x^{2}+ 9\) is a factor.

The easiest way to approach this problem is to break everything into exponents. \(\displaystyle 8^{3}\) is equal to \(\displaystyle 2^{9}\) and 27 is equal to \(\displaystyle 3^{3}\). Therefore, the expression can be broken down into \(\displaystyle \frac{2^{9} \times 3}{2^{6}\times 3^{3}}\). When you cancel out all the terms, you get \(\displaystyle \frac{2^{3}}{3^{2}}\), which equals \(\displaystyle \frac{8}{9}\).