Question

Simplify -

$\tt \blue{(2a - 3b + 3c)^{2} - (2a - 3b - 4c) ^{2} }$

​

Answer 1

Step-by-step explanation:

(x+y+z)

2

=x

2

+y

2

+z

2

+2xy+2yz+2zx [Identity]

Here,

x=2a

y=3b

z=4c

Putting Values in Identity, We get :

⇒ (2a+3b+4c)

2

=(2a)

2

+(3b)

2

+(4c)

2

+2(2a)(3b)+2(3b)(4c)+2(4c)(2a)

=4a

2

+9b

2

+16c

2

+12ab+24bc+16ca

Answer 2

ANSWER:

To Simplify:

• (2a - 3b + 3c)² - (2a - 3b - 4c)²

Solution:

We are given that,

⇒ (2a - 3b + 3c)² - (2a - 3b - 4c)²

We know that,

⇒ (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Here in first term,

⇒ x = 2a, y = (-3b) and z = 3c

And in second term,

⇒ x = 2a, y = (-3b) and z = (-4c)

So,

⇒ (2a - 3b + 3c)² - (2a - 3b - 4c)²

⇒ [(2a)² + (-3b)² + (3c)² + 2((2a)(-3b) + (-3b)(3c) + (3c)(2a))] - [(2a)² + (-3b)² + (-4c)² + 2((2a)(-3b) + (-3b)(-4c) + (-4c)(2a))]

⇒ [4a² + 9b² + 9c² + 2((-6ab) + (-9bc) + (6ca))] - [4a² + 9b² + 16c² + 2((-6ab) + (12bc) + (-8ca))]

So,

⇒ [4a² + 9b² + 9c² + 2(-6ab - 9bc + 6ca)] - [4a² + 9b² + 16c² + 2(-6ab + 12bc - 8ca)]

⇒ [4a² + 9b² + 9c² - 12ab - 18bc + 12ca] - [4a² + 9b² + 16c² - 12ab + 24bc - 16ca]

Opening the bracket,

⇒ 4a² + 9b² + 9c² - 12ab - 18bc + 12ca - 4a² - 9b² - 16c² + 12ab - 24bc + 16ca

Grouping like terms together,

⇒ (4a² - 4a²) + (9b² - 9b²) + (9c² - 16c²) + (-12ab + 12ab) + (-18bc - 24bc) + (12ca + 16ca)

So,

⇒ (0) + (0) + (-7c²) + (0) + (-44bc) + (28ca)

Hence,

⇒ - 7c² - 44bc + 28ca

Taking c common,

⇒ c(-7c - 44b + 28a)

Therefore,

⇒ (2a - 3b + 3c)² - (2a - 3b + 4c)² = c(-7c - 44b + 28a)

OR

⇒ (2a - 3b + 3c)² - (2a - 3b + 4c)² = c(28a - 44b - 7c)

Formula Used:

• (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)