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in Linear Inequations by (46.3k points)
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The value of 64a3 + 48a2b + 12a2b+b3 at a= 1 and b = -1 is

(a) 25 

(b) 125 

(c) 27 

(d) 54

1 Answer

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Best answer

(c) 27

64a3 + 48a2b + 12a2b + b3

⇒ (4a + b)3   [∴ (a + b)2 = a3 + b3 + 3a2b + 3ab2]

∴ Reqd. value = (4-1)3 = 33 = 27.

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