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Question
If a, b, c is in A.P., prove that:
a3 + c3 + 6abc = 8b3.
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Solution
Since a, b, c are in A.P., we have:
2b = a+c
\[\Rightarrow\] b = \[\frac{a + c}{2}\]
Consider RHS:
8 \[b^3\]
\[\text { Substituting b } = \frac{a + c}{2}: \]
\[ \Rightarrow 8 \left( \frac{a + c}{2} \right)^3 \]
\[ \Rightarrow a^3 + c^3 + 3ac\left( a + c \right)\]
\[ \Rightarrow a^3 + c^3 + 3ac(2b)\]
\[ \Rightarrow a^3 + c^3 + 6abc\]
Hence, proved.
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