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 \(\displaystyle y=\frac{4}{5}x-\frac{27}{5}\)