Question
If a, b, c are in A.P., prove that:
$\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
$\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
✓
Answer
If $\text{a}^3+\text{c}^3+6\text{abc}=8\text{c}^3$
or $\text{a}^3+\text{c}^3-(2\text{b})^3+6\text{abc}=0$
or $\text{a}^3+(-2\text{b})^3+\text{c}^3+3\times\text{a}\times(-2\text{b})\times\text{c}=0$
$\therefore(\text{a}-2\text{b}+\text{c})=0$ $\begin{bmatrix}\therefore\text{x}^3+\text{y}^3+\text{z}^3+3\text{xyz}=0\\\text{or if}\ \text{x}+\text{y}+\text{z}=0\end{bmatrix}$
or $\text{a}+\text{c}=2\text{b}$
$\text{a}-\text{b}=\text{c}-\text{b}$
and since, a, b, c are in A.P
Thus, $\text{a}-\text{b}=\text{c}-\text{d}$
Hence proved. $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
or $\text{a}^3+\text{c}^3-(2\text{b})^3+6\text{abc}=0$
or $\text{a}^3+(-2\text{b})^3+\text{c}^3+3\times\text{a}\times(-2\text{b})\times\text{c}=0$
$\therefore(\text{a}-2\text{b}+\text{c})=0$ $\begin{bmatrix}\therefore\text{x}^3+\text{y}^3+\text{z}^3+3\text{xyz}=0\\\text{or if}\ \text{x}+\text{y}+\text{z}=0\end{bmatrix}$
or $\text{a}+\text{c}=2\text{b}$
$\text{a}-\text{b}=\text{c}-\text{b}$
and since, a, b, c are in A.P
Thus, $\text{a}-\text{b}=\text{c}-\text{d}$
Hence proved. $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
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