simplify the following:
(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-3b)
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Answer:
We solve the given expression (a−2b+5c)(a−b)−(a−b−c)(2a+3c)+(6a+b)(2c−3a−5b)
(a−2b+5c)(a−b)−(a−b−c)(2a+3c)+(6a+b)(2c−3a−5b)
=[(a−b)(a−2b+5c)]−[(2a+3c)(a−b−c)]+[(6a+b)(2c−3a−5b)]
=[a(a−2b+5c)−b(a−2b+5c)]−[2a(a−b−c)+3c(a−b−c)]+[6a(2c−3a−5b)+b(2c−3a−5b)]
=(a
2
−2ab+5ac−ab+2b
2
−5bc)−(2a
2
−2ab−2ac+3ac−3bc−3c
2
)+(12ac−18a
2
−30ab+2bc−3ab−5b
2
)
=(a
2
+2b
2
−3ab+5ac−5bc)−(2a
2
−3c
2
−2ab+ac−3bc)+(−18a
2
−5b
2
+12ac−33ab+2bc)
=−19a
2
−3b
2
−3c
2
−38ab+18ac−6bc
Therefore, (a−2b+5c)(a−b)−(a−b−c)(2a+3c)+(6a+b)(2c−3a−5b)=−19a
2
−3b
2
−3c
2
−38ab+18ac−6bc
Hence, (a−2b+5c)(a−b)−(a−b−c)(2a+3c)+(6a+b)(2c−3a−5b)=19ca−37ab−19a
2
is false.
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