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Answers
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) + (6bc) + (-4ca)
Hence,
⇒ - 7c² + 6bc - 4ca
Taking c common,
⇒ c(-7c + 6b - 4a)
Therefore,
⇒ (2a - 3b + 3c)² - (2a - 3b + 4c)² = c(-7c + 6b - 4a)
OR
⇒ (2a - 3b + 3c)² - (2a - 3b + 4c)² = c(-4a + 6b - 7c)
Formula Used:
- (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
Answer:
(2a - 3b + 3c)² - (2a - 3b + 4c)² = -7c²-4ac+6bc
Step-by-step explanation:
Here,
Let
(2a - 3b + 3c) = x
and,
(2a - 3b + 4c) = y
We know that,
x²-y² = (x+y) (x-y)
Substituting the values,
(2a - 3b + 3c)² - (2a - 3b + 4c)²
= {(2a-3b+3c) + (2a-3b+4c)} {(2a-3b+3c) - (2a-3b+4c)}
Solving the brackets,
= {2a-3b+3c+2a-3b+4c} {2a-3b+3c-2a+3b-4c}
Rearranging like terms,
= (2a+2a-3b-3b+3c+4c) (2a-2a-3b+3b+3c-4c)
= (4a-6b+7c)(-c)
= -c(4a-6b+7c)
Multiplying the terms,
= -4ac+6bc-7c²
Rearranging the terms,
= -7c²-4ac+6bc
Hence,
(2a - 3b + 3c)² - (2a - 3b + 4c)² = -7c²-4ac+6bc