Three numbers form an arithmetic progression (A.P). The sum of the numbers is equal to $3$ and the sum of their cubes is equal to $4$. Find the numbers.
Let the first term of the A.P be $= a$ $\;$ and the common difference $= d$
Let the three terms of A.P be $\;$ $a - d, \; a, \; a + d$
Given: $\;$ Sum of the three numbers is equal to $3$
i.e. $\;$ $a -d + a + a + d = 3$
i.e. $\;$ $3a = 3$ $\implies$ $a = 1$
And: $\;$ Sum of the cubes of these three numbers is equal to $4$
i.e. $\;$ $\left(a - d\right)^3 + a^3 + \left(a + d\right)^3 = 4$
i.e. $\;$ $\left(1 - d\right)^3 + 1^3 + \left(1 + d\right)^3 = 4$ $\;\;\;$ $\left[\because \; a = 1\right]$
i.e. $\;$ $\left(1 - d + 1 + d\right) \left[\left(1 - d\right)^2 - \left(1 - d\right) \left(1 + d\right) + \left(1 + d\right)^2\right] = 3$
$\left[\because \; p^3 + q^3 = \left(p + q\right) \left(p^2 - pq + q^2\right)\right]$
i.e. $\;$ $2 \left[1 - 2d + d^2 - 1 + d^2 + 1 + 2d + d^2\right] = 3$
i.e. $\;$ $3 d^2 + 1 = \dfrac{3}{2}$
i.e. $\;$ $3 d^2 = \dfrac{1}{2}$
i.e. $\;$ $d^2 = \dfrac{1}{6}$ $\implies$ $d = \pm \dfrac{1}{\sqrt{6}}$
When $\;$ $a = 1$ $\;$ and $\;$ $d = \dfrac{+ 1}{\sqrt{6}}$, $\;$ the three numbers are $\;$ $1 - \dfrac{1}{\sqrt{6}}$, $\;$ $1$, $\;$ $1 + \dfrac{1}{\sqrt{6}}$
When $\;$ $a = 1$ $\;$ and $\;$ $d = \dfrac{- 1}{\sqrt{6}}$, $\;$ the three numbers are $\;$ $1 + \dfrac{1}{\sqrt{6}}$, $\;$ $1$, $\;$ $1 - \dfrac{1}{\sqrt{6}}$
$\therefore \;$ The three numbers in A.P are $\;$ $1 + \dfrac{1}{\sqrt{6}}, \; 1, \; 1 - \dfrac{1}{\sqrt{6}}$