Algebra
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If the sum of a and its reciprocal is 1 and a ≠ 0, b ≠ 0, then the value of a3 + b3 is b
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According to question,
a + b = 1 a a
⇒ a2 + b2 = ab
⇒ a2 – ab + b2 = 0
∴ a3 + b3 = (a + b)( a2 – ab + b2 ) = 0
Correct Option: C
According to question,
a + b = 1 a a
⇒ a2 + b2 = ab
⇒ a2 – ab + b2 = 0
∴ a3 + b3 = (a + b)( a2 – ab + b2 ) = 0
- If a = √7 + 2√12 and b = √7 - 2√12 , then ( a3 + b3 ) is equal to
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a = √7 + 2 × √4 × √3
= √4 + 3 + 2 × 2 × √3
= √( 2 + √3 )2 = 2 + √3
∴ b = √7 - 2√12 = 2 - √3
⇒ a + b = 2 + √3 + 2 - √3 = 4
ab = (2 + √3)(2 - √3) = 1
∴ a3 + b3 = ( a + b )3 - 3ab(a + b)
= 64 – 3 × 4 = 52Correct Option: D
a = √7 + 2 × √4 × √3
= √4 + 3 + 2 × 2 × √3
= √( 2 + √3 )2 = 2 + √3
∴ b = √7 - 2√12 = 2 - √3
⇒ a + b = 2 + √3 + 2 - √3 = 4
ab = (2 + √3)(2 - √3) = 1
∴ a3 + b3 = ( a + b )3 - 3ab(a + b)
= 64 – 3 × 4 = 52
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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
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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
- 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