find a number where the sum of itself and its cube is the same as twice the square if it
Let x be the number.
"the sum of itself and its cube is the same as twice the square if it"
x+x^3 = 2x^2.
Solve by factoring
x^3-2x^2+x=0
x(x^2-2x+1)=0
x(x-1)^2=0
x = 0,1
To find a number where the sum of itself and its cube is the same as twice the square of it, we can represent this as a mathematical equation and solve for the number.
Let's denote the number as "x." The sum of the number itself and its cube is x + x^3. On the other hand, twice the square of the number is 2x^2.
So, we have the equation x + x^3 = 2x^2.
To solve this equation, we can follow these steps:
1. Rewrite the equation in the form of a polynomial equation: x^3 + x - 2x^2 = 0.
2. Rearrange the terms to get the equation in standard polynomial form: x^3 - 2x^2 + x = 0.
3. Factor out an x: x(x^2 - 2x + 1) = 0.
4. Factor the quadratic expression inside the parentheses: x(x - 1)^2 = 0.
Now, we have two potential solutions: x = 0 and x = 1. Let's check if they satisfy the original equation.
For x = 0: 0 + 0^3 = 0, and twice the square of 0 is also 0. So, this solution works.
For x = 1: 1 + 1^3 = 2, but twice the square of 1 is also 2. Therefore, this solution is valid as well.
So, we found two numbers that satisfy the given condition - 0 and 1.