By the perfect square trinomial pattern,

$\displaystyle (x+ y) ^{2}$

$\displaystyle =x^{2} +2xy+ y ^{2}$

$\displaystyle x^{2} = 8$ and $\displaystyle y^{2} = 18$.

Also, by the Power of a Power Principle, 

$\displaystyle (xy) ^{2} = x^{2} y^{2} = 8 \cdot 18 = 144$, 

so, since $\displaystyle x$ and $\displaystyle y$ are both positive, 

$\displaystyle xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$.

Therefore, 

$\displaystyle (x+ y) ^{2}$

$\displaystyle =x^{2} +2xy+ y ^{2}$

$\displaystyle = 8 +2 (12)+18$

$\displaystyle = 8 +24 +18$

$\displaystyle = 50$

By the Power of a Power Principle, 

$\displaystyle (y ^{3} ) ^{3} = y ^{3 \cdot 3} = y ^{9}$

Therefore, we substitute, keeping in mind that an odd power of a negative number is also negative:

$\displaystyle y ^{9} = (y ^{3} ) ^{3}$

$\displaystyle = (-17) ^{3}$

$\displaystyle = - 17 ^{3}$

$\displaystyle = - (17 \cdot 17\cdot 17)$

$\displaystyle = -4,913$

By the perfect square trinomial pattern,

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

$\displaystyle =x^{2} -2xy+ y ^{2}$

$\displaystyle x^{2} =1 8$ and $\displaystyle y^{2} = 8$.

Also, by the Power of a Power Principle, 

$\displaystyle (xy) ^{2} = x^{2} y^{2} = 18 \cdot 8 = 144$, 

so, since $\displaystyle x$ and $\displaystyle y$ are both positive,

$\displaystyle xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$.

Therefore, 

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

$\displaystyle =x^{2} -2xy+ y ^{2}$

And, substituting:

$\displaystyle x^{2} -2xy+ y ^{2}$

$\displaystyle =18 -2 (12)+8$

$\displaystyle =18 -24 +8$

$\displaystyle = 2$

Multiply out the expression by using multiple distributions and collecting like terms:

$\displaystyle (t-x) (t^{2}+tx+x^{2})$

$\displaystyle = t (t^{2}+tx+x^{2}) - x (t^{2}+tx+x^{2})$

$\displaystyle = t \cdot t^{2}+ t \cdot tx+ t \cdot x^{2} - x \cdot t^{2}- x \cdot tx- x \cdot x^{2}$

$\displaystyle = t^{3}+ t^{2} x+ t x^{2} - t^{2} x- tx^{2} - x^{3}$

$\displaystyle = t^{3}+ t^{2} x- t^{2} x + t x^{2} - tx^{2} - x^{3}$

$\displaystyle = t^{3} - x^{3}$

Since $\displaystyle (t^{3} ) ^{2} = t ^{3 \cdot 2 } = t^{6}$ by the Power of a Power Principle,

$\displaystyle t^{3} = \pm \sqrt{ t^{6}} = \pm \sqrt{144} = \pm 12$.

However, $\displaystyle t$ is positive, so $\displaystyle t^{3}$ is as well, so we choose $\displaystyle t^{3} = 12$.

Similarly, 

$\displaystyle x^{3} = \pm \sqrt{x^{6}} = \pm \sqrt{81} = \pm 9$.

However, since $\displaystyle x$ is negative, as an odd power of a negative number, $\displaystyle x^{3}$ is as well, so we choose $\displaystyle x^{3} = -9$.

Therefore, substituting:

$\displaystyle (t-x) (t^{2}+tx+x^{2}) = t^{3} - x^{3} = 12 - (-9) = 21$ 

If $\displaystyle B$ is odd, then $\displaystyle A + B +1$, the sum of three odd integers, is odd; an odd number taken to any positive integer power is odd.

If $\displaystyle B$ is even, then $\displaystyle A + B +1$, the sum of two odd integers and an even integer, is even; an even number taken to any positive integer power is even. 

Therefore, $\displaystyle (A + B + 1)^{A+B+1}$ always assumes the same odd/even parity as $\displaystyle B$.

We can use the point slope form of a line, substituting $\displaystyle m = 1, x_{1} = 4, y _{1} = 7$.

$\displaystyle y- y_{1} = m (x-x_{1})$

$\displaystyle y- 7 = 1 (x-4)$

$\displaystyle y- 7 = x-4$

$\displaystyle y- 7 -y +4 = x-4 -y +4$

$\displaystyle -3 = x-y$

or 

$\displaystyle x-y = -3$

You will first solve for Y, to get the equation in $\displaystyle y=mx+b$ form.

$\displaystyle m$ represents the slope of the line, which would be $\displaystyle -\frac{2}{3}$.

A perpendicular line's slope would be the negative reciprocal of that value, which is $\displaystyle \frac{3}{2}$.

First, find the slope of the line.

$\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

Now, because the problem tells us that the line goes through $\displaystyle (0,2)$, our y-intercept must be $\displaystyle 2$.

Putting the pieces together, we get the following equation:

$\displaystyle y=\frac{2}{3}x+2$

To find the equation of a line, we need to first find the slope.

$\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-6}{-2-(-4)}=\frac{-8}{2}=-4$

Now, our equation for the line looks like the following:

$\displaystyle y=-4x+b$

To find the y-intercept, plug in one of the given points and solve for $\displaystyle b$. Using $\displaystyle (-4, 6)$, we get the following equation:

$\displaystyle 6=-4(-4)+b$

Solve for $\displaystyle b$.

$\displaystyle 6=16+b$

$\displaystyle b=-10$

Now, plug the value for $\displaystyle b$ into the equation.

$\displaystyle y=-4x-10$

First, we need to find the slope of the line.

$\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-1}{-2-8}=\frac{-8}{-10}=\frac{4}{5}$

Next, find the $\displaystyle y$-intercept. To find the $\displaystyle y$-intercept, plug in the values of one point into the equation $\displaystyle y=mx+b$, where $\displaystyle m$ is the slope that we just found and $\displaystyle b$ is the $\displaystyle y$-intercept.

$\displaystyle 1=8(\frac{4}{5})+b$

Solve for $\displaystyle b$.

$\displaystyle 1=\frac{32}{5}+b$

$\displaystyle b=-\frac{27}{5}$

Now, put the slope and $\displaystyle y$-intercept together to get $\displaystyle y=\frac{4}{5}x-\frac{27}{5}$