If 3x+2y=12 and xy=6,find 27x^3+8y^3
27x^3+8y^3 --- the difference of cubes
= (3x + 2y)(9x^2 - 6xy + 4y^2)
= 12(9x^2 - 6xy + 4y^2)
but 9x^2 - 6xy + 4y^2
= (9x^2 + 12xy + 4y^2) - 18xy
= (3x+2y)^2 - 18xy
= 144 - 108
= 36
so 27x^3 + 8y^3 = 12(36) = 4320
or:
y = 6/x
back in the first:
3x + 2(6/x) = 12
times x
3x^2 + 12 = 12x
3x^2 - 12x + 12 = 0
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2
y = 6/2 = 3
then 27x^3+8y^3
= 27(8) + 8(27) = 432
Simple one
To find the value of 27x^3 + 8y^3, we need to substitute the given values of x and y into the equation.
Let's solve the given equations:
Equation 1: 3x + 2y = 12
Equation 2: xy = 6
We can solve Equation 1 for x:
3x = 12 - 2y
x = (12 - 2y) / 3
x = (4 - (2/3)y)
Now, substitute the value of x into Equation 2:
(4 - (2/3)y) * y = 6
Expand the equation:
4y - (2/3)y^2 = 6
Multiply the equation by 3 to remove the fraction:
12y - 2y^2 = 18
Rearrange the equation to get it in the standard form:
2y^2 - 12y + 18 = 0
We now have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.
Let's solve it using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = -12, and c = 18.
y = (-(-12) ± √((-12)^2 - 4(2)(18))) / (2(2))
y = (12 ± √(144 - 144)) / 4
y = (12 ± √0) / 4
Since the discriminant (b^2 - 4ac) is zero, we have only one real solution for y.
y = 12 / 4
y = 3
Now that we have the value of y, we can substitute it back into Equation 1 to find the value of x:
3x + 2(3) = 12
3x + 6 = 12
3x = 12 - 6
3x = 6
x = 2
Now that we have the values of x = 2 and y = 3, we can substitute them into the expression 27x^3 + 8y^3:
27(2^3) + 8(3^3)
27(8) + 8(27)
216 + 216
432
Therefore, the value of 27x^3 + 8y^3 is 432.