$a^3+b^3+c^3=d^3$

Algebra Level 2

Find the set of four positive consecutive integers such that the sum of the cubes of the first three is equal to the cube of the fourth.

Suppose $a , b, c, \text {and } d$ are the integers obtained respectively, then what is $a+b+c+d=?$

18 23 17 22 20

1 solution

Tom Engelsman
Feb 12, 2018

We require (for $x \in \mathbb{N}$ ):

$x^3 + (x+1)^3 + (x+2)^3 = (x+3)^3 \Rightarrow x^3 - 6x - 9 = 0 \Rightarrow (x-3)(x^2 + 3x + 3) = 0$

which yields $x = 3, \frac{-3 \pm \sqrt{3}i}{2}$ for the roots. Hence, the required consecutive positive integers are $3, 4, 5, 6$ which sum to $\boxed{18}.$

Thank you for sharing a nice solution.

Hana Wehbi - 3 years, 3 months ago

a 3 + b 3 + c 3 = d 3 a^3+b^3+c^3=d^3 a 3 + b 3 + c 3 = d 3

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$a^3+b^3+c^3=d^3$