Question

evaluate. (x+y) (2x+y2+3x) (x2+y2-1) ,at x= 1, y= -1​

Answer 1

Step-by-step explanation:

(1+(-1))(2(1)+(-1)2+3(1))(1(2)+(-1)2-1)

0(2-2+3)(2-2-1)

(2-5)(-1)

(-3)(-1)

3

Answer 2

QUESTION: Evaluate. (x+y) (2x+y2+3x) (x2+y2-1) ,at x= 1, y= -1

GIVEN: x = 1 & y = (-1)

SOLUTION: Putting the value of x & y in the equation, we get:

$= > (x + y)( 2x + {y}^{2} + 3x)( {x}^{2} + {y}^{2} - 1)$

$= > (1 - 1)(2 \times 1 + { (- 1)}^{2} + 3 \times 1)( {(1)}^{2} + {( - 1)}^{2} - 1)$

$= > (0)(2 + 1 + 3)(1 + 1 - 1) \\ = > (0)(6)(1) \\ = > 0$

Hence, The answer is 0