Question

Need answer from this pic

Answer 1

Let the three numbers be x,x + 1,x + 2.

Given that Sum of cubes of three numbers is 36.

⇒ x³ + (x + 1)³ + (x + 2)³ = 36

⇒ x³ + x³ + 3x² + 3x + 1 + x³ + 6x² + 12x + 8 = 36

⇒ 3x³ + 9x² + 15x - 27 = 0

⇒ 3(x³ + 3x² + 5x - 9) = 0

⇒ x³ + 3x² + 5x - 9 = 0

⇒ x³ + 4x² - x² + 9x - 4x - 9 = 0

⇒ x³ + 4x² + 9x - x² - 4x - 9 = 0

⇒ x(x² + 4x + 9) - 1(x² + 4x + 9) = 0

⇒ (x - 1)(x² + 4x + 9) = 0

⇒ x = 1, Neglect x^2 + 4x + 9.

⇒ x = 1.

Then:

⇒ x + 2 = 3

⇒ x + 3 = 4.

Therefore, the numbers are 1,3,4.

Hope this helps!