Question

If a, b, c is in A.P., prove that:

 a³ + c³ + 6abc = 8b³.

Answer 1

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.