Group and combine like terms \(\displaystyle 5x^{2} y ^{2},6x^{2}y^{2}\):

\(\displaystyle 5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy\)

\(\displaystyle = 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7\)

\(\displaystyle =\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7\)

\(\displaystyle =11x^{2}y^{2}+ 11 xy +7\)

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

Since \(\displaystyle x\) and \(\displaystyle y\) have different signs,

\(\displaystyle xy< 0\), and, subsequently,

\(\displaystyle 2xy < 0\)

Therefore, 

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

This makes (b) the greater quantity.

Simplify the expression in (a):

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

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

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

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

Therefore, whether (a) or (b) is greater depends on the values of \(\displaystyle x\) and \(\displaystyle y\), neither of which are known. 

We give at least one positive value of \(\displaystyle x\) for which (a) is greater and at least one positive value of \(\displaystyle x\) for which (b) is greater.

Case 1: \(\displaystyle x = 2\)

(a) \(\displaystyle x^{2} + 3x = 2^{2} + 3 \cdot 2 = 4 + 6 = 10\)

(b) \(\displaystyle 4x = 4 \cdot 2 = 8\)

Case 2: \(\displaystyle x = \frac{1}{2}\)

(a) \(\displaystyle x^{2} + 3x = \left ( \frac{1}{2} \right ) ^{2} + 3 \cdot \frac{1}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}= 1 \frac{3}{4}\)

(b) \(\displaystyle 4x = 4 \cdot \frac{1}{2} = 2\)

Therefore, either (a) or (b) can be greater.

Any nonzero expression raised to the power of 0 is equal to 1. Therefore, 

\(\displaystyle \left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0} = 1 + 1 = 2\).

None of the given expressions are correct.

\(\displaystyle \left (\frac{x^{3}}{2y^{4}} \right ) ^{-3} = \left (\frac{2y^{4}}{x^{3}} \right ) ^{3} = \frac{2^{3} \left (y^{4}\right ) ^{3} }{\left (x^{3} \right )^{3}}= \frac{8 y^{4 \cdot 3} }{x^{3\cdot 3} }= \frac{8 y^{12} }{x^{9} }\)

If \(\displaystyle y > 1\), then \(\displaystyle y ^{2}> 1^{2} = 1\) and \(\displaystyle y ^{-2}= \frac{1}{y ^{2}} < \frac{1}{1} = 1\)

\(\displaystyle y ^{2}> 1> y ^{-2}\), so by transitivity, \(\displaystyle y ^{2}> y ^{-2}\), and (b) is greater

By the Binomial Theorem, if \(\displaystyle (x + 1)^{n}\) is expanded, the coefficient of \(\displaystyle x^{r}\) is

 \(\displaystyle _{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}\).

(a) Substitute \(\displaystyle n = 10, r = 3\): The coerfficient of \(\displaystyle x^{3}\) is 

\(\displaystyle _{10}\textrm{C} _{3} = \frac{10!}{(10-3)!3!} = \frac{10!}{7!3!}\).

(b) Substitute \(\displaystyle n = 10, r = 7\): The coerfficient of \(\displaystyle x^{7}\) is 

\(\displaystyle _{10}\textrm{C} _{7} = \frac{10!}{(10-7)!7!} = \frac{10!}{3!7!} = \frac{10!}{7!3!}\).

The two are equal.

\(\displaystyle \left (-2 \right )^{-3} = \left ( -\frac{1}{2} \right )^{3}\)

A negative number to an odd power is negative, so the expression in (a) is negative. The expression in (b) is positive since the base is positive. (b) is greater.

Simplify the expression in (a):

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

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

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

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

\(\displaystyle =30 x^{2} + 250\)

Since \(\displaystyle x>0\), 

\(\displaystyle (x + 5)^{3}- (x - 5)^{3}=30 x^{2} + 250 > 250\),

making (a) greater.