Solve -1000+y^3, where y is a real number

y^3-1000

ty

what do you mean by "solve"?

It can be factored as the difference of two cubes:
(y-10)(y^2+10y+100)

If you want the solutions to
y^2-1000 = 0
then that is just y=10 or y=-5(1±√3 i)

To solve the expression -1000 + y^3, where y is a real number, we need to find the value of y which satisfies the equation.

Since the equation does not have an equal sign, we cannot solve for a specific value of y. However, we can simplify the expression by factorizing it if possible.

To simplify the expression, we can start by factoring out a common factor of -1 from -1000 + y^3:

-1000 + y^3 = -1(1000 - y^3)

Now, we can see that we have a difference of cubes expression within the parentheses - (1000 - y^3). We can apply the formula for difference of cubes, which states:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

In our case, a = 10 and b = y:

1000 - y^3 = (10)^3 - y^3
= (10 - y)(10^2 + 10y + y^2)

So, now our simplified expression becomes:

-1000 + y^3 = -1((10 - y)(10^2 + 10y + y^2))

Therefore, the simplified expression for -1000 + y^3, where y is a real number, is -1((10 - y)(10^2 + 10y + y^2)).