Question

If a 1, b=0 and e=-1, find the values
of the following expressions
0 (a2-3ac +a-3) (b-a-b2 -
2ab)​

Answer 1

Answer:

$- 2$

Step-by-step explanation:

$a = 1$

$b = 0$

$c = - 1$

$({a}^{2} - 3ac + a - 3)(b - a - {b}^{2} - 2ab)$

putting the given values

$( {1}^{2}- 3(1)( - 1) +1 - 3)(0 - 1 - {0}^{2} - 2(1)(0) \: \: )$

$(1 - ( - 3) - 2)( - 1)$

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

$(2)( - 1)$

$- 2$

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