Exponents Equation

Question

In the equation w^3+x^3+y^3=z^3 where w^3,x^3,y^3, and z^3 are distinct consecutive positive perfect cubes listed in ascending order. What is the smallest possible value of z?

Note: A list of table is ok

Answer 1

By condition w = x -1, y = x + 1 and z = x + 2. We get equatiopn (x - 1)³ + x³ + (x + 1)³ = (x + 2)³;

x³ - 3x² + 3x - 1 + x³ + x³ + 3x² + 3x + 1 = x³ + 6x² + 12x + 8; after simplification we have: 2x³ - 6x² - 6x - 8 = 0, or x³ - 3x² - 3x - 4 = 0. Using rational zero theorem we have x = 4 is zero. Checking using synthetic division:

4 | 1 -3 -3 -4

4 4 4

1 1 1 0

Setting quotion x² + x + 1 = 0 we get quadratic equation with discriminant 1² - 4·1·1 = - 3 < 0.

It means that we have x = 4 is uniq. Then this numbers: w = 3, x = 4, y = 5 and z = 6.

Answer: z = 6 and uniq

Answer 2

The first three positive perfect cubes are 1, 8, and 27, and they are the result of 1^3, 2^3, and 3^3, respectively; they are listed in a consecutive and ascending order. Because of this, if w^3, x^3, and y^3 are summed together, the result would be 36 = z^3. 36 isn't a perfect positive cube, but we can still take the square root of 36 in order to get the value of z. The cubed root of 36 is 3.30193..., which is the smallest possible value of z.

I hope this explanation is clear, thank you.