Question

If 2x + 3y = 12 and xy = 6, find the value of 8x^3 + 27y^3​

Answer 1

Answer:

The value of $8\cdot x^{3} + 27\cdot y^{3}$ is 432.

Step-by-step explanation:

Let be the following system of equations:

$2\cdot x + 3\cdot y = 12$ (1)

$x\cdot y = 6$ (2)

Then, we solve both for $x$ and $y$:

From (1):

$2\cdot x + 3\cdot y = 12$

$2\cdot x = 12- 3\cdot y$

$x = 6 - \frac{3}{2}\cdot y$

(1) in (2):

$\left(6-\frac{3}{2}\cdot y \right)\cdot y = 6$

$6\cdot y-\frac{3}{2}\cdot y^{2} = 6$

$\frac{3}{2}\cdot y^{2}-6\cdot y + 6 = 0$

The roots of the polynomial are determined by the Quadratic Formula:

$y_{1} = y_{2} = 2$

By (1):

$x = 6 - \frac{3}{2}\cdot (2)$

$x = 3$

If we know that $x = 3$ and $y = 2$, then the final value is:

$z = 8\cdot x^{3}+27\cdot y^{3}$

$z = 8\cdot 3^{3}+27\cdot 2^{3}$

$z = 432$

The value of $8\cdot x^{3} + 27\cdot y^{3}$ is 432.