Question

x³+3x²y+3xy²+y³

if (x = -2. y=-3)​

Answer 1

Answer:

Rearrange the equation to (x – y)³ + 3x²y – 3xy² = -11³ or

(y – x) [ (y – x)² + 3xy ] = 11³

Since 11 is a prime integer, the only possible values for y – x are 1, 11, 11² or 11³

If y – x = 1 then 3xy = 11³ -1 = 1330 or xy = 1330/3 which is not an integer and therefore this is not a possible scenario.

If y – x = 11 then 3xy = 0 and either x or y must be 0 and not positive as required.

If y – x = 11² then 3xy = 11 - 11² or xy <0 and either x or y must be negative and not both being positive as required.

If y – x = 11³ then 3xy = 1 - 11³ or xy <0 which is the same as the previous case.

Step-by-step explanation:

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Answer 2

Answer:

putting x = -2

and

y = -3

-2³ + 3×(-2)²(-3) + 3 (-2) (-3) ² + (-3) ³

= - 8 + (-36) + (-36) + (-27)

= -8 - 36 - 36 - 27

= -107