Question

if the sum of three numbers in AP is and sum of their cubes is 288. find the number.​

Answer 1

Answer:

Hera 3a=12

Hera 3a=12a=4 

Hera 3a=12a=4 Also

$({a - d})^{3} + {a}^{3} + ( {a + d})^{3} = 288$

or

${3a}^{3} + {6ad}^{2} = 288$

${24d}^{2} = 288 - 3 \times 64 = 96$

${d}^{2} = 4$

d = ±2

Hence the numbers are 2,4,6 or 6,4,2

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Answer 2

Answer:

the number are 2,4,6 or 6,4,2

Step-by-step explanation:

let a-b,a+d be the no. in AP

thus,a-d+a+a+d=12

=3a=12

a=4

also we have,

(a-d)^e + a^3+(a+d)^3=288

3a^3+24d^2=288

3×64+24d^2 = 288

192+24d^2=288

24d^2= 288-192

24d^2=96

d^2=4

d=2

: thus the no. are :4-2,4,4+2 or 4-(-2)4,4+(-2)

the no.are :2,4,6 or 6,4,2