Question

Using the identity (a+b)? = a² +6² +2ab, derine the formula for (a + b + c)². Hence find the value of (22-3y + 4z)?​

Answer 1

Answer:

Let (a+b) = x and c = y --------- {1}

Therefore (a+b+c)^2 can be written as (x + y)^2

On expanding we get

x^2 + y^2 + 2xy --------- {2}

Now placing the value of {1} in {2} we get

(a+b)^2 + c^2 + 2*(a+b)*c

=> (a^2 + b^2 +2ab) + c^2 + 2ac + 2bc

=> a^2 + b^2 + c^2 + 2ab + 2bc +2ca

=> a^2 + b^2 + c^2 + 2(ab + bc + ca) ---------- {3}

In the equation given (22 - 3y + 4z)^2

a = 22, b = -3y and c = 4z

Now we place the value of a, b and c in {3} we get

(22)^2 + (-3y)^2 + (4z)^2 + 2 [ (22*(-3y)) + ((-3y)*4z) + (22*4z)]

=> 484 + 9y^2 + 16z^2 + 2 [ -66y -12yz + 88z]

=> 484 + 9y^2 + 16z^2 + [ -132y -24yz + 176z]

=> 484 + 9y^2 + 16z^2 - 132y - 24yz + 176z {ANS}