Question

a=3,b=1 then find 4(a^2+b^2+2ab)-[4(a^2+b^2-2ab)-{-b^3+4(a-3)}] find solutions​

Answer 1

Answer is 71

Step-by-step explanation:

Given,

$4[a^2 + b^2 + 2ab)] - [4(a^2 + b^2 - 2ab) - (-b^3 + 4(a - 3))]$

⇒ $4[a^2 + b^2 + 2ab)] - [4a^2 + 4b^2 - 8ab) - (-b^3 + 4a - 12))]$

⇒ $4[a^2 + b^2 + 2ab)] - [4(a^2 + b^2 - 2ab) - (-b^3 + 4(a - 3))]$

⇒ $16ab - b^3 + 4a + 12$

Now, Putting the values of a = 3 and b = 1, we get

$(-1) + 16\times (3)\times (1) + 4\times(3) + 12$

= -1 + 48 + 12 + 12

= 71

Answer is 71