Question

Find the product of (3x+y) and (x²-y).​

Answer 1

Answer:

given

   x = y + 2

     x² + y² = 34

   x - y = 2

 squaring,  (x - y)² = 2²

           x² + y² - 2 x y = 4

           34 - 2 x y = 4

             x y = 15

===========================

2x - 3y = 10          x y  = 16

8x³ - 27 y³ = (2x - 3y) ( 4x² + 6 xy + 9 y²)

             = (2 x - 3y) [ (2x -3y)² + 18 xy ]

             = 10 * [ 10² + 18 * 16 ]

====================================

x > 0  and  x² + 1/9x²  = 25 / 36

    (x + 1/3x)²  = x² + 1/9x² + 2/3  = 25/36  + 2/3

           (x + 1/3x)² = 49/36

               x + 1/3x = 7/6      , we drop the negative value as x > 0.

 x³ +  1/27x³  = (x +  1/3x) ( x² + 1/9x² - x/3x)

                   = 7/6 * (25/36 - 1/3)

                   = 7/6 * 1/36

                    = 7/216

Step-by-step explanation:

Answer 2

Answer:

(3 X+Y) [(X)2-Y)]

3$x^{3}$ - 3$\\ XY$ + $x^{2}$Y - $Y^{2}$

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