Solving the System of Equations: Finding the Values of $a^2 - ab + b^2$ and $a^2 + ab + b^2$
Given the system of equations:
$a + b = 10 \tag{1}$ $ab = 16 \tag{2}$
We are asked to find the values of $a^2 - ab + b^2$ and $a^2 + ab + b^2$.
Step 1: Solve for $a$ and $b$
From equation $(1)$, we can express $b$ in terms of $a$:
$b = 10 - a \tag{3}$
Substitute equation $(3)$ into equation $(2)$:
$a(10 - a) = 16$
Expand and simplify:
$10a - a^2 = 16$
Rearrange to form a quadratic equation:
$a^2 - 10a + 16 = 0$
Factor the quadratic:
$(a - 8)(a - 2) = 0$
This gives us two possible values for $a$:
$a = 2, 8$
Substitute these values back into equation $(3)$ to find the corresponding values of $b$:
$b = 10 - 2 = 8, \quad b = 10 - 8 = 2$
So, we have two possible solutions: $(a, b) = (2, 8)$ or $(a, b) = (8, 2)$.
Step 2: Calculate $a^2 - ab + b^2$ and $a^2 + ab + b^2$
Using the solutions from Step 1, we can calculate the required expressions:
Case 1: $(a, b) = (2, 8)$
$a^2 - ab + b^2 = 2^2 - 2(8) + 8^2 = 4 - 16 + 64 = \boxed{52}$
$a^2 + ab + b^2 = 2^2 + 2(8) + 8^2 = 4 + 16 + 64 = \boxed{84}$
Case 2: $(a, b) = (8, 2)$
$a^2 - ab + b^2 = 8^2 - 8(2) + 2^2 = 64 - 16 + 4 = \boxed{52}$
$a^2 + ab + b^2 = 8^2 + 8(2) + 2^2 = 64 + 16 + 4 = \boxed{84}$
As we can see, both cases yield the same results:
$a^2 - ab + b^2 = \boxed{52}$ $a^2 + ab + b^2 = \boxed{84}$
