Solve -1000+y^3, where y is a real number
Simplify your answer as much as possible.
If there is more than one solution, separate them with commas.
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)
ty
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)).