If the mode and the variance of the numbers $a, b, 8, 5$ and 10 are 6 and 8 respectively, then the value of $a^3 + b^3$ is equal to
Step-by-Step Solution
Key Concept: Variance and mean constraints together with sum of squares provide a system to solve for unknown values.
Given mean $= 6$ and variance $= 6.8$. From the mean: $\frac{a+b+c+1+0}{5} = 6$, so $a + b + c = 29$. From variance $6.8 = \frac{\sum x_i^2}{5} - 36$, we get $\sum x_i^2 = 234$. We have $a^2 + b^2 + c^2 + 1 = 234$, so $a^2 + b^2 + c^2 = 233$. Using $2ab = (a+b)^2 - a^2 - b^2$ and the constraint $a + b + c = 29$: Let $a^2 + b^2 = 25$, then $c = 29 - a - b$ and $c^2 = 208$. Solving: $2ab = (a+b)^2 - 25$, and from $(a+b)^2 + (a-b)^2 = 2(a^2+b^2) = 50$ with other constraints, we find $a = 7, b = 3, c = 19$ (or similar). Thus $(a+b)^2 + (a-b)^2 = 3ab + 12 = 91$.
Correct Answer: 3