Question

What are the real zeros of

y = (x + 3)^3 + 10?

Answer 1

using the sum of two cubes,

(x + 3)^3 + 10
= (x+3+∛10)((x+3)^2 - (x+3)∛10 + ∛100)
= x^2 + (6-∛10)x + (∛100-3∛10+9)

One root is clearly x = -(3+∛10).

You can see that the discriminant of the quadratic is negative, so there are no other real roots.

Or, consider the graph. It is just the graph of y=x^3, shifted left 3 and up 10. You know it crosses the x-axis in only one place, so you know there is only one real root, which you have found.