If 3x+2y=12 and xy=6,find 27x^3+8y^3
write it in form of a^3+b^3 as (3x)^3 + (2x)^3 now apply identity and open it
27x^3+8y^3 = (3x+2y)(9x^2-6xy+4y^2)
(3x+2y)^2 = 144 = 9x^2+12xy+4y^2
. . .
done yesterday
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why did you repeat the question?
To find the value of 27x^3 + 8y^3, we can use the given equations to find the values of x and y.
Let's start by solving the system of equations:
1) 3x + 2y = 12
2) xy = 6
First, let's solve Equation 2) for y:
xy = 6
Divide both sides by x:
y = 6/x
Now, substitute this value of y into Equation 1):
3x + 2(6/x) = 12
Multiply both sides by x to eliminate the fraction:
3x^2 + 12 = 12x
Rearrange the equation:
3x^2 - 12x + 12 = 0
Divide the entire equation by 3 to simplify:
x^2 - 4x + 4 = 0
Factor the quadratic equation:
(x - 2)^2 = 0
Take the square root of both sides:
x - 2 = 0
x = 2
Now that we have found the value of x, we can substitute it back into Equation 2) to find y:
xy = 6
2y = 6
Divide both sides by 2:
y = 3
So, we have found that x = 2 and y = 3.
Now, let's substitute these values into the expression 27x^3 + 8y^3:
27(2)^3 + 8(3)^3
Simplify the exponents:
27(8) + 8(27)
Multiply:
216 + 216
Add:
432
Therefore, 27x^3 + 8y^3 = 432 when x = 2 and y = 3.