Q. 285.0( 1 Vote )

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Answer :

Given that: (4a βˆ’ b + 2c) 2


= (4a) 2 + (βˆ’b) 2 + (2c)2 + 2(4a)(βˆ’b) + 2(βˆ’b)(2c) + 2(2c)(4a)


[∡ (a + b + c) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)]


= 16a2 + b2 + 4c2 – 8ab – 4bc + 16ca


(ii). Given that: (3π‘Ž βˆ’ 5𝑏 βˆ’ 𝑐)2


= (3a)2 + (βˆ’5b)2 + (βˆ’c)2 + 2(3a)(βˆ’5b) + 2(βˆ’5b)(βˆ’c) + 2(βˆ’c)(3a)


[∡ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)]


= 9a2 + 25b2 + c2 – 30ab + 10bc – 6ca


(iii). Given that: (βˆ’π‘₯ + 2𝑦 βˆ’ 3𝑧)2


= (βˆ’π‘₯)2 + (2𝑦)2 + (βˆ’3𝑧)2 + 2(βˆ’π‘₯)(2𝑦) + 2(2𝑦)(βˆ’3𝑧) + 2(βˆ’3𝑧)(βˆ’π‘₯)


[∡ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)]


= x2 + y2 + 9z2 – 4xy – 12yz + 6zx


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