The sum of 3 numbers is 4, the sum of their product is 5
Find the sum of their cube if the product of there number is 6
I can make no sense at all of your problem statement.
Hmmm. Maybe you mean
x+y+z = 4
xy + xz + yz = 5
xyz = 6
Since that has no real solutions, I don't think that's what you meant.
To find the sum of the cubes of the three numbers, we first need to find the three numbers themselves. From the given information, we know that the sum of the three numbers is 4 and the sum of their product is 5.
Let's assume the three numbers are a, b, and c. We can write the given information as equations:
Equation 1: a + b + c = 4
Equation 2: ab + ac + bc = 5
Next, we need to find the values of a, b, and c. To do this, we'll use a technique called algebraic manipulation.
We'll start by squaring equation 1 to eliminate the brackets:
(a + b + c)^2 = (4)^2
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 16
Now, let's simplify the equation:
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 16
We can rearrange this equation by subtracting equation 2 (ab + ac + bc = 5) from it:
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - (ab + ac + bc) = 16 - 5
a^2 + b^2 + c^2 + ab + ac + bc = 11
Now, we have the sum of the squares of the three numbers (a^2 + b^2 + c^2) in terms of their products and their sum. We'll denote this as equation 3.
To find the product of the three numbers (abc), we'll multiply equation 2 by equation 1:
(ab + ac + bc) * (a + b + c) = (5) * (4)
a^2b + a^2c + ab^2 + abc + ac^2 + b^2c + bc^2 = 20
Now, we have the sum of the products of the three numbers (a^2b + a^2c + ab^2 + abc + ac^2 + b^2c + bc^2) in terms of their product. We'll denote this as equation 4.
Given that the product of the three numbers (abc) is 6, we can substitute this value into equation 4:
a^2b + a^2c + ab^2 + abc + ac^2 + b^2c + bc^2 = 20
a^2b + a^2c + ab^2 + 6 + ac^2 + b^2c + bc^2 = 20
Simplifying the equation, we get:
a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 = 14
Now, we have the sum of the products of the three numbers (a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2) in terms of their sum. We'll denote this as equation 5.
To find the sum of the cubes of the three numbers (a^3 + b^3 + c^3), we'll multiply equation 5 by equation 1:
(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2) * (a + b + c) = (14) * (4)
a^3 + a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 = 56
Substituting the value of abc as 6 (from earlier), we get:
a^3 + a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 = 56
a^3 + 14 = 56
Finally, solving for a^3, we get:
a^3 = 56 - 14
a^3 = 42
Therefore, the sum of the cubes of the three numbers is 42.