Question

The sum of three numbers in ap is 12 and the sum of their cubes is 288. find the numbers.

Answer 1

let the number be a,a+d,a-d
a=4,d=2
so the number are 2,4,6

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{p\;be\;the\;first\;term}}$

Also

$\textbf{\underline{d\;be\;the\;common\;difference}}$

Three numbers are :-

p - d , p , p + d

Situation :-

p - d + p + p + d = 12

3p = 12

$\tt{\rightarrow p=\dfrac{12}{3}}$

= 4

Hence we get :-

p = 4

Now

${\boxed{\sf\:{2^nd \;Situation}}}$

(p - d)³ + (p)³ + (p + d)³ = 288

(4 - d)³ + (4)³ + (4 + d)³ = 288

(4 - d)³ + 64 + (4 + d)³ = 288

(4 - d)³ + (4 + d)³ = 288 - 64

(4 - d)³ + (4 + d)³ = 224

65 - d³ - 48d + 12d² + 65 + d³ + 48d + 12d² = 224

24d² = 224 - 64 - 64

24d² = 224 - 128

24d² = 96

${\boxed{\sf\:{d^2=\dfrac{96}{24}}}}$

d² = 4

d = √4

d = ±2

Hence numbers are :-

4 - 2 , 4 , 4 + 2 or 4 + 2 , 4 , 4 - 2

$\Large{\boxed{\sf\:{2,4,6\;or\;6,4,2}}}$