Question:

Step 1: Simplify the first logarithmic equation. Given the equation $\log\left(\dfrac{8b^3 - a^3 - c^3}{6abc}\right) = 0$. For the logarithm to be defined, $\dfrac{8b^3 - a^3 - c^3}{6abc} > 0$. Since $\log x = 0$ implies $x=1$, we have: $$ \dfrac{8b^3 - a^3 - c^3}{6abc} = 1 $$ $$ 8b^3 - a^3 - c^3 = 6abc $$ $$ 8b^3 = a^3 + c^3 + 6abc \quad (*)$$ Step 2: Simplify the second logarithmic equation and establish relationships between $a, b, c$. Given the equation $\log b = \log(a^2 - 4) = \log(c - 2)$. For the logarithms to be defined, $b > 0$, $a^2 - 4 > 0$, and $c - 2 > 0$. From the equality of logarithms, we deduce: $$ b = a^2 - 4 \quad (1) $$ $$ b = c - 2 \quad (2) $$ From equation (2), we express $c$ in terms of $b$: $$ c = b + 2 $$ From equation (1), we express $a^2$ in terms of $b$: $$ a^2 = b + 4 $$ Step 3: Solve for $a, b, c$. Substitute $c = b+2$ and $a^2 = b+4$ into equation $(*)$: $$ 8b^3 = a^3 + (b+2)^3 + 6ab(b+2) $$ We seek integer values for $a, b, c$. From $a^2 = b+4$, if $a$ is an integer, $b+4$ must be a perfect square. Let's test integer values for $a$. If $a=3$, then $a^2=9$. From $a^2 = b+4$, we get $9 = b+4$, which implies $b=5$. Now, using $c=b+2$, we get $c=5+2=7$. Let's verify these values $a=3, b=5, c=7$ with the original conditions: 1. Logarithm domain conditions: $b=5 > 0$ (satisfied) $a^2-4 = 3^2-4 = 9-4=5 > 0$ (satisfied) $c-2 = 7-2=5 > 0$ (satisfied) 2. Equation $(*)$: $8b^3 = 8(5^3) = 8(125) = 1000$. $a^3 + c^3 + 6abc = 3^3 + 7^3 + 6(3)(5)(7) = 27 + 343 + 630 = 370 + 630 = 1000$. The values $a=3, b=5, c=7$ satisfy all given conditions. Step 4: Determine the sequence type and general term. The first three terms of the sequence are $T_1 = a = 3$, $T_2 = b = 5$, $T_3 = c = 7$. This is an arithmetic progression (AP) with first term $T_1 = 3$ and common difference $d = T_2 - T_1 = 5 - 3 = 2$. The $n^{\text{th}}$ term of this AP is given by $T_n = T_1 + (n-1)d$: $$ T_n = 3 + (n-1)2 = 3 + 2n - 2 = 2n+1 $$ The sum of the first $n$ terms of this AP is given by $S_n = \dfrac{n}{2}(2T_1 + (n-1)d)$: $$ S_n = \dfrac{n}{2}(2(3) + (n-1)2) = \dfrac{n}{2}(6 + 2n - 2) = \dfrac{n}{2}(2n+4) = n(n+2) $$ Step 5: Calculate the required values. Using $T_n = 2n+1$ and $S_n = n(n+2)$: 1. Calculate $T_{10}$: $$ T_{10} = 2(10) + 1 = 20 + 1 = 21 $$ 2. Calculate $S_{10}$: $$ S_{10} = 10(10+2) = 10(12) = 120 $$ 3. Calculate $T_{21}^2 - 2T_{21}T_1 + 2T_1^2$: First, find $T_1$ and $T_{21}$: $T_1 = 3$. $T_{21} = 2(21) + 1 = 42 + 1 = 43$. Now substitute these values into the expression: $$ T_{21}^2 - 2T_{21}T_1 + 2T_1^2 = (43)^2 - 2(43)(3) + 2(3)^2 $$ $$ = 1849 - 258 + 18 $$ $$ = 1591 + 18 = 1609 $$ 4. Calculate $S_{11} - S_{10}$: The difference $S_{11} - S_{10}$ represents the $11^{\text{th}}$ term of the sequence, $T_{11}$. $$ T_{11} = 2(11) + 1 = 22 + 1 = 23 $$ Alternatively, calculate $S_{11}$ and subtract $S_{10}$: $S_{11} = 11(11+2) = 11(13) = 143$. $S_{11} - S_{10} = 143 - 120 = 23$.