Question

Expand
(1/2a-1/3b+c)2

Answer 1

Using identity - (a+b+c)^2= a^2 + b^2 +c^2 +2ab +2bc + 2ca

We get ,
(1/2a)^2+ (-1/3b)^2+ (c)^2 +{ 2(1/2a)(-1/3b)} +{2 (-1/3b)(c) }+ {2 (c) (1/2a)}

= a - 2/3b + c^2 -1/3ab -2/3bc +ac

Answer 2

Given:

$(1/2a-1/3b+c)^{2}$

To find:

The expanded form of the given expression

Solution:

The expanded form  $(1/2a-1/3b+c)^{2}$ is $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ +(3bc-2ac-1)/3ab.

We can expand by following the given steps-

We know that the expression can be expanded using the following identity-

$(x+y+z)^{2}$= $x^{2} +y^{2} +z^{2} +2(xy+yz+zx)$

On comparing with the given expression, x=1/2a, y=(-1)/3b, z=z.

Using these values, we get

$(1/2a-1/3b+c)^{2}$= $(1/2a)^{2}$+$(-1/3b)^{2}$ +$c^{2}$ +2(1/2a×(-1)/3b+(-1)/3b×c+c×1/2a)

=$1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ +2(-1/6ab-c/3b+c/2a)

= $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ -1/3ab-2c/3b+c/a

Now we will take the LCM of the last 3 terms and add them.

= $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ -c/3abc-2a$c^{2}$/3abc+3b$c^{2}$/3abc

= $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ +(3b$c^{2}$-2a$c^{2}$-c)/3abc

= $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ +(3bc-2ac-1)/3ab

Therefore, the expanded form  $(1/2a-1/3b+c)^{2}$ is $1/4a^{2}$+ $1/9b^{2}$ +$c^{2}$ +(3bc-2ac-1)/3ab.