+2 To find the value of $a - b$, we can subtract the second equation from the first equation:
\begin{align*}
(a^3 + 3ab^2) - (a^3 - 3ab^2) &= 679 - 615 \
6ab^2 &= 64 \
ab^2 &= \frac{64}{6} \
ab^2 &= \frac{32}{3}
\end{align*}
Now, we can substitute this value of $ab^2$ into one of the original equations. Let's use the first equation:
\begin{align*}
a^3 + 3ab^2 &= 679 \
a^3 + 3\left(\frac{32}{3}\right) &= 679 \
a^3 + 32 &= 679 \
a^3 &= 679 - 32 \
a^3 &= 647
\end{align*}
Now, we can take the cube root of both sides to solve for $a$:
\begin{align*}
a &= \sqrt[3]{647} \
a &= 7
\end{align*}
Finally, we can substitute the value of $a$ back into one of the original equations to find the value of $b$. Let's use the first equation again:
\begin{align*}
7^3 + 3(7)b^2 &= 679 \
343 + 21b^2 &= 679 \
21b^2 &= 679 - 343 \
21b^2 &= 336 \
b^2 &= \frac{336}{21} \
b^2 &= 16 \
b &= \pm 4
\end{align*}
Since we are interested in finding $a - b$, we take the positive value of $b$:
\begin{align*}
a - b &= 7 - 4 \
a - b &= 3
\end{align*}
Therefore, $a - b = \boxed{3}$. MyBKExperience Survey Code
+6 To find the area of triangle ABC with AB = 6, BC = 8, and angle ABC = 90 degrees, we can use the formula for the area of a right triangle:
Area = (1/2) * base * height
In this case, AB is the base and BC is the height:
Area = (1/2) * AB * BC
= (1/2) * 6 * 8
= 24 square units.
Therefore, the area of triangle ABC is 24 square units.
Now, let's find angle BAC when angle ABC = 135 degrees and angle ACB = 30 degrees.
The sum of the angles in a triangle is always 180 degrees. So, we can find angle BAC by subtracting the sum of angles ABC and ACB from 180 degrees: Angle BAC = 180 degrees - Angle ABC - Angle ACB
= 180 degrees - 135 degrees - 30 degrees
= 180 degrees - 165 degrees
= 15 degrees. MyBKExperience Survey
Therefore, angle BAC is 15 degrees.
