Question

Three numbers form an arithmetic progression. The sum of the numbers is equal to $3$ and the sum of their cubes is equal to $4$. Find the numbers.

Answer 1

Let the $3$ numbers be $a-d,a\&a+d$

$\therefore a-d+a+a+d=3\Rightarrow 3a=3\Rightarrow a=1-\left(i\right)$

Sum of cubes

$\Rightarrow {(a-d)}^3+a^3+{(a+d)}^3=4\Rightarrow a^3-3a^{2}b+3a{d}^2-d^3+a^3+a^3+3a^{2}d+3a{d}^2+d^3=4\Rightarrow {3a}^3+6a{d}^2=4$

Replacing $a=1$ from $\left(i\right)$

$\Rightarrow 3\times 1^2+6\times 1\times d^2=4\Rightarrow 3+6d^2=4\Rightarrow 6d^2=1\Rightarrow d^2=\frac{1}{6}\Rightarrow d=\frac{\pm 1}{\sqrt{6}}$

The numbers are

$1-\left(\frac{\pm 1}{\sqrt{6}}\right),1,1+\left(\frac{\pm 1}{\sqrt{6}}\right)$