If three unequal numbers  a, b, c  are in A.P. and  b − a, c − b , a are in G.P., then  a3 + b3 + c3abc  is equal to

The problem states that three unequal numbers a, b, c are in Arithmetic Progression (A.P.) and b - a, c - b, a are in Geometric Progression (G.P.). We are asked to find the value of a³ + b³ + c³ - 3abc.

Step 1: Since a, b, c are in A.P., we have 2b = a + c.

Step 2: Since b - a, c - b, a are in G.P., we have (c - b)² = a(b - a). Substituting 2b = a + c from step 1, we get (2b - b)² = a(b - b/2), which simplifies to b² = ab/2, or 2b = a.

Step 3: Now, we know that a = 2b and c = 2b. Substituting these values into the expression a³ + b³ + c³ - 3abc, we get (2b)³ + b³ + (2b)³ - 3(2b)(b)(2b).

Step 4: Simplifying the above expression, we get 8b³ + b³ + 8b³ - 12b³ = 5b³.

So, a³ + b³ + c³ - 3abc = 5b³.