If the three terms \[a,b,c\] are in A.P., then prove that the terms \[{a^2}\left( {b + c} \right),{b^2}\left( {c + a} \right){\text{ and }}{c^2}\left( {a + b} \right)\] are also in A.P.

Hint: First of all, find the condition in which the first three terms are in A.P. Then let us assume that the second three terms are in A.P. and state the condition to be in A.P. Then prove that by using the obtained equations to reach the solution of the given problem.

Complete step by step solution:
Given that the terms \[a,b,c\] are in A.P.
We know that if the three terms \[x,y,z\] are in A.P. then they should satisfy the condition that \[2y = x + z\].
So, for the terms \[a,b,c\] which are in A.P., we have
\[2b = a + c............................................\left( 1 \right)\]
Now, we have to prove that \[{a^2}\left( {b + c} \right),{b^2}\left( {c + a} \right){\text{ and }}{c^2}\left( {a + b} \right)\] are in A.P.
Let us assume that these three terms are in A.P.
So, they should satisfy that \[2{b^2}\left( {a + c} \right) = {a^2}\left( {b + c} \right) + {c^2}\left( {a + b} \right)\]
Grouping the common terms, we have
\[
   \Rightarrow 2{b^2}\left( {a + c} \right) = {a^2}b + {a^2}c + a{c^2} + b{c^2} \\
   \Rightarrow 2{b^2}\left( {a + c} \right) = b\left( {{a^2} + {c^2}} \right) + ac\left( {c + a} \right) \\
\]
Adding and subtracting \[2ac\] in the brackets of \[{a^2} + {c^2}\], we have
\[
   \Rightarrow 2{b^2}\left( {a + c} \right) = b\left( {{a^2} + {c^2} + 2ac - 2ac} \right) + ac\left( {c + a} \right) \\
   \Rightarrow 2{b^2}\left( {a + c} \right) = b\left( {{a^2} + 2ac + {c^2}} \right) - 2abc + ac\left( {c + a} \right) \\
   \Rightarrow 2{b^2}\left( {a + c} \right) = b{\left( {a + c} \right)^2} - 2abc + ac\left( {c + a} \right){\text{ }}\left[ {\because \left( {{a^2} + 2ac + {c^2}} \right) = {{\left( {a + c} \right)}^2}} \right] \\
\]
Substituting \[c + a = 2b\] from equation \[\left( 1 \right)\], we have
\[
   \Rightarrow 2{b^2}\left( {2b} \right) = b{\left( {2b} \right)^2} - 2abc + 2abc \\
  \therefore 4{b^3} = 4{b^3} \\
\]
Hence, our assumption is correct.
Thus, we have proved that the terms \[{a^2}\left( {b + c} \right),{b^2}\left( {c + a} \right){\text{ and }}{c^2}\left( {a + b} \right)\] are in A.P.

Note: A sequence of numbers is called an Arithmetic progression (A.P.) if the difference between any two consecutive terms is always the same. If the three terms \[x,y,z\] are in A.P. then they should satisfy the condition that \[2y = x + z\].