\(\displaystyle \left (4x \right )^{4} +\left ( 2y \right )^{2} =4^{4}x^{4} + 2^{2}y^{2}=256x^{4} +4y ^{2}\)

A binomial can be cubed using the pattern:

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

Set \(\displaystyle A = x, B = 5\)

\(\displaystyle (x-5)^{3}\)

\(\displaystyle = x^{3} - 3x^{2} \cdot 5+ 3x\cdot 5^{2} - 5^{3}\)

\(\displaystyle = x^{3} - 15x^{2} + 75x - 125\)

A trinomial whose leading term has a coefficent other than 1 can be factored using the \(\displaystyle ac\)-method. We split the middle term using two numbers whose product is \(\displaystyle 3 (-14) = -42\) and whose sum is \(\displaystyle -1\). These numbers are \(\displaystyle -7,6\), so:

\(\displaystyle 3x^{2} - x -14\)

\(\displaystyle = 3x^{2} - 7x +6x -14\)

\(\displaystyle = \left (3x^{2} - 7x \right ) +\left ( 6x -14 \right )\)

\(\displaystyle =x \left (3x- 7 \right ) +2 \left (3x- 7 \right )\)

\(\displaystyle =\left ( x+2 \right ) \left (3x- 7 \right )\)

This can be achieved by using the pattern of difference of squares:

\(\displaystyle \left ( x + y + 3 \right ) \left ( x + y - 3\right )\)

\(\displaystyle =\left [\left ( x + y \right ) + 3 \right ] \left [\left ( x + y \right ) - 3 \right ]\)

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

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

Applying the binomial square pattern:

\(\displaystyle = x^{2} + 2xy + y^{2} - 9\)

The cube of a sum pattern can be applied here:

\(\displaystyle \left (x + \frac{1}{3} \right )^{3}\)

\(\displaystyle = x ^{3 } + 3\cdot x^{2} \cdot \frac{1}{3}+3\cdot x \cdot \left ( \frac{1}{3} \right )^{2}+\left( \frac{1}{3} \right )^{3}\)

\(\displaystyle = x ^{3 } + x^{2} +3\cdot x \cdot \frac{1}{9} +\frac{1}{27}\)

\(\displaystyle = x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}\)

To obtain the constant term of a perfect square trinomial, divide the linear coefficient, which here is \(\displaystyle -13\), by 2, and square the quotient. The result is 

\(\displaystyle \left (\frac{ -13}{2} \right) ^{2} = \frac{(-13)^{2}}{2^{2}}= \frac{169}{4}\)

To obtain the constant term of a perfect square trinomial, divide the linear coefficient, which here is \(\displaystyle -18\), by 2, and square the quotient. The result is 

\(\displaystyle \left (\frac{-18}{2} \right )^{2} =\left ( -9\right ) ^{2} = 81\)

Using the Binomial Theorem, if \(\displaystyle (x + 2)^{n}\) is expanded, the \(\displaystyle x^{r}\) term is 

\(\displaystyle _{n}\textrm{C} _{r} \cdot x^{r} \cdot 2 ^{n-r}\)

\(\displaystyle = \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r} \cdot x^{r}\).

This makes \(\displaystyle \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r}\) the coefficient of \(\displaystyle x^{r}\).

We compare the values of this expression at \(\displaystyle n = 10\) for both \(\displaystyle r = 4\) and \(\displaystyle r = 6\):

(a) \(\displaystyle \frac{10!}{(10-4)!4!} \cdot 2 ^{10-4} = \frac{10!}{6!4!} \cdot 2 ^{6}\)

(b) \(\displaystyle \frac{10!}{(10-6)!6!} \cdot 2 ^{10-6} = \frac{10!}{4!6!} \cdot 2 ^{4} = \frac{10!}{6!4!} \cdot 2 ^{4}\)

(a) is the greater quantity.

\(\displaystyle 16x^{2} - 72x + 81\) 

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

\(\displaystyle =\left ( 4 x \right ) ^{2} - 2 \cdot 4x \cdot 9 + 9^{2}\)

\(\displaystyle = \left ( 4x -9 \right )^{2}\)

Since \(\displaystyle x > 3\), 

\(\displaystyle 4x - 9 > 4 \cdot3 - 9 = 3\), so

\(\displaystyle 16x^{2} - 72x + 81= \left ( 4x -9 \right )^{2} > 3^{2} = 9 >8\)

making (a) greater. 

We show that either polynomial can be greater by giving two cases:

Case 1: \(\displaystyle y = 0\)

\(\displaystyle y ^{2} + 4y + 4 = 0 ^{2} + 4 \cdot 0 + 4 = 4\)

\(\displaystyle y^{2} + 6y + 9 = 0^{2} + 6 \cdot 0 + 9 = 9\)

\(\displaystyle y^{2} + 6y + 9 > y ^{2} + 4y + 4\)

Case 2: \(\displaystyle y = -3\)

\(\displaystyle y ^{2} + 4y + 4 = (-3) ^{2} + 4 \cdot (-3) + 4 = 9-12 + 4 = 1\)

\(\displaystyle y^{2} + 6y + 9 = (-3)^{2} + 6 \cdot (-3) + 9 = 9 - 18 + 9 = 0\)

\(\displaystyle y ^{2} + 4y + 4 < y^{2} + 6y + 9\)