1. The problem presents the function $$t = \frac{3 w^3 + a}{w^3 - 2}$$ and provides a curve passing through points $(-2,8)$ and $(2,-2)$ on the coordinate plane. 2. We will use the given points to find any unknowns and verify the function behavior. 3. Substitute $w = -2$ and $t=8$ into the function: $$8 = \frac{3(-2)^3 + a}{(-2)^3 - 2} = \frac{3(-8) + a}{-8 - 2} = \frac{-24 + a}{-10}$$ Multiply both sides by $-10$: $$8 \times (-10) = -24 + a \implies -80 = -24 + a$$ Solve for $a$: $$a = -80 + 24 = -56$$ 4. Substitute $w = 2$ and $t = -2$ to verify value of $a$: $$-2 = \frac{3(2)^3 + a}{(2)^3 - 2} = \frac{3(8) + a}{8 - 2} = \frac{24 + a}{6}$$ Multiply both sides by $6$: $$-12 = 24 + a$$ Solve for $a$: $$a = -12 - 24 = -36$$ 5. From steps 3 and 4, two different values of $a$ are obtained ($-56$ and $-36$), meaning the function $$t = \frac{3 w^3 + a}{w^3 - 2}$$ cannot pass through both points simultaneously with a constant $a$. 6. Given this contradiction, the function as defined cannot model the curve passing through both points simultaneously without adjusting $a$ for each. Final answer: There is no single value of $a$ such that $$t = \frac{3 w^3 + a}{w^3 - 2}$$ passes through both $(-2,8)$ and $(2,-2)$.