Question

use suitable identity and expand ​

Answer 1

Answer: (a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + [(-3b+a)/2]

Step-by-step explanation:

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

x = a/4, y = -b/2, z = 1

(a/4 -b/2 + 1)^2 = (a/4)^2 + (-b/2)^2 + (1)^2 + 2[(a/4 * -b/a) + (-b/2 * 1) + (a/4 * 1) ]

(a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + 2[(-b/4) + (-b/2) + (a/4)]

(a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + 2[(-b - 2b/4) + (a/4)]

(a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + 2[(-3b/4) + a/4]

(a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + 2[(-3b+a)/4]

(a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + [(-3b+a)/2]

Answer: (a/4 -b/2 + 1)^2 = a^2/16 + b^2/4 + 1 + [(-3b+a)/2]

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