Algebra
- If P = 999, then the value of ³√P( P² + 3P + 3 ) + 1 is
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P = 999 (Given)
Now , ³√P( P² + 3P + 3 ) + 1
= ³√( P³ + 3P² + 3P + 1 )
= ³√( P + 1 )³
= P + 1 = 999 + 1 = 1000Correct Option: A
P = 999 (Given)
Now , ³√P( P² + 3P + 3 ) + 1
= ³√( P³ + 3P² + 3P + 1 )
= ³√( P + 1 )³
= P + 1 = 999 + 1 = 1000
- If a = 4 . 36, b = 2.39 and c = 1.97, then the value of a3 - b3 - c3 – 3abc is
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Using Rule 21,
Here, a – b – c = 4.36 – 2.39 – 1.97 = 0 ( ∴ a - b - c = 0)
∴ a3 - b3 - c3 = 3abc = 0
⇒ a3 - b3 - c3 - 3abc = 0Correct Option: C
Using Rule 21,
Here, a – b – c = 4.36 – 2.39 – 1.97 = 0 ( ∴ a - b - c = 0)
∴ a3 - b3 - c3 = 3abc = 0
⇒ a3 - b3 - c3 - 3abc = 0
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x + 1 
x - 1 x2 + 1 - 1 x2 + 1 + 1 is equal to x x x2 x2
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∴ x + 1 x - 1 x2 + 1 - 1 x2 + 1 + 1 x x x2 x2 = 
x2 - 1 x2 + 1 2 - 1 
x2 x2 = x2 - 1 x4 + 1 + 1 x2 x4 Required answer = x6 - 1 x6
Correct Option: D
∴ x + 1 x - 1 x2 + 1 - 1 x2 + 1 + 1 x x x2 x2 = x2 - 1 x2 + 1 2 - 1 x2 x2 = x2 - 1 x4 + 1 + 1 x2 x4 Required answer = x6 - 1 x6
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If a = 11 and b = 9, then the value of a2 + b2 + ab is : a3 - b3
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a2 + ab + b2 = a2 + ab + b2 a3 - b3 ( a - b )( a2 + ab + b2 ) = 1 a - b = 1 = 1 11 - 9 2 Correct Option: A
a2 + ab + b2 = a2 + ab + b2 a3 - b3 ( a - b )( a2 + ab + b2 ) = 1 a - b = 1 = 1 11 - 9 2
- If x2 + y2 + 1 = 2x , then the value of x3 + y5 is
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x2 + y2 + 1 = 2x
⇒ x2 + y2 + 1 - 2x = 0
⇒ x2 - 2x + 1 + y2 = 0
⇒ (x – 1)2 + y2 = 0
⇒ x – 1 = 0
⇒ x = 1 and y = 0
∴ x3 + y5 = 1 + 0 = 1Correct Option: D
x2 + y2 + 1 = 2x
⇒ x2 + y2 + 1 - 2x = 0
⇒ x2 - 2x + 1 + y2 = 0
⇒ (x – 1)2 + y2 = 0
⇒ x – 1 = 0
⇒ x = 1 and y = 0
∴ x3 + y5 = 1 + 0 = 1
