The grouping technique works here:

$\displaystyle t^{3} +4t^{2}-9t -36$

$\displaystyle =\left (t^{3} +4t^{2} \right )-\left (9t +36 \right )$

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

$\displaystyle =t ^{2} \left (t + 4} \right )-9 \left ( t + 4 \right )$

$\displaystyle =\left (t ^{2}-9 \right ) \left (t + 4} \right )$

The first factor is the difference of squares and can be factored further accordingly:

$\displaystyle \left (t ^{2}-9 \right ) \left (t + 4} \right )$

$\displaystyle =\left (t ^{2}- 3^{2} \right ) \left (t + 4} \right )$

$\displaystyle =\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )$

Since the first term is a perfect cube, the factoring pattern we are looking to take advantage of is the difference of cubes pattern. However, 243 is not a perfect cube of an integer $\displaystyle (6^{3} = 216 < 243 < 7^{3} = 343)$, so the factoring pattern cannot be applied.  No other pattern fits, so the polynomial is a prime.

$\displaystyle \left (x^{2} + x + 7 \right )^{2} = \left (x^{2} + x + 7 \right )\left (x^{2} + x + 7 \right )$

Vertical multiplication is perhaps the easiest way to multiply trinomials.

                         $\displaystyle x^{2} + x + 7$

                         $\displaystyle \underline{x^{2} + x + 7}$

                     $\displaystyle 7x^{2} + 7x + 49$

              $\displaystyle x ^{3 } + x^{2} + 7x$

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

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

The difference of two terms can be cubed using the pattern

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

Where $\displaystyle A = 9R, B = 10$:

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

$\displaystyle (9R-10)^{3} = (9R) ^{3}-3(9R)^{2}\cdot 10 + 3(9R) \cdot 10^{2}-10^{3}$

$\displaystyle (9R-10)^{3} =729R^{3}-3\cdot 81R^{2}\cdot 10 + 3(9R) \cdot 100-1,000$

$\displaystyle (9R-10)^{3} =729R^{3}-2,430R^{2} + 2,700R-1,000$

Factor $\displaystyle y^{4}- 81$ all the way to its prime factorization.

$\displaystyle y^{4}- 81$ can be factored as the difference of two perfect square terms as follows:

$\displaystyle y^{4}- 81$

$\displaystyle =\left ( 9y^{2} \right )^{2} - 1^{2}$

$\displaystyle =\left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )$

$\displaystyle 9y^{2} + 1$ is a factor, and, as the sum of squares, it is a prime. $\displaystyle 9y^{2} - 1$ is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

$\displaystyle \left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )$

$\displaystyle =\left ( 9y^{2} + 1 \right )\left [ \left (3y \right ) ^{2} - 1^{2} \right ]$

$\displaystyle =\left ( 9y^{2} + 1 \right )\left (3y+1 \right )\left (3 y-1\right )$

Therefore, of the given four choices, only $\displaystyle 3y^{2} + 1$ is not a factor, so the correct response is three.

First, factor the numerator. We need factors that multiply to $\displaystyle \small -6$ and add to $\displaystyle \small -5$.

$\displaystyle \small 1*-6=-6\ \text{and}\ 1+(-6)=-5$

$\displaystyle \small x^2-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\displaystyle \small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}$

Note that $\displaystyle \small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$\displaystyle \frac{(x-6)(x+1)}{(x+1)}=(x-6)$

This is our final answer.

When simplifying an equation,you must find a common factor for all values in the equation, including both sides.  

$\displaystyle 3,18,$and, $\displaystyle 81$ can all be divided by $\displaystyle 3$ so divide them all at once 

$\displaystyle \left ( \frac{3}{3}=1, \frac{18}{3}=6, \frac{81}{3}=27\right )$.  

This leaves you with 

$\displaystyle x+6y=27$.

$\displaystyle \frac{x^5(3x^2y^3)}{x^4y^2z}$

Simplify the numerator by multiplying in the $\displaystyle \tiny x^5$ term

$\displaystyle \frac{3x^7y^3}{x^4y^2z}$

Cancel out like terms in the numerator and denominator.

$\displaystyle \frac{3x^3y}{z}$

Rewrite the denominator of the second fraction using a power.

$\displaystyle (\frac{x^{-1}}{x}) = (\frac{x^{-1}}{x^1})$

Using the subtraction rule of exponents, we can simplify this as one term.

$\displaystyle x^{-1-1} = x^{-2}$

The expression becomes:

$\displaystyle (x^2)\cdot (x^{-2})$

Apply the addition rule of exponents.

$\displaystyle x^{2+(-2)} = x^0 =1$

The answer is:  $\displaystyle 1$

Factor an $\displaystyle x^6$ in the numerator to get:

$\displaystyle \frac{(x^7 - x^6)}{ (x - 1)} =\frac{x^6(x-1)}{(x-1)}$

We can now cancel out $\displaystyle (x-1)$ from the numerator and denominator leaving the answer.

$\displaystyle x^6$