Algebra
- If a + b = 1, then a4 + b4 – a3 – b3 – 2a2b2 + ab is equal to
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a4 + b4 – a3 – b3 – 2a2b2 + ab
= a4 + b4 – 2a2b2 – a3 – b3 + ab
= (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
= (a – b)2 – a2 + ab – b2 + ab
[∵ a + b = 1]
= (a – b)2 – (a – b)2 = 0Correct Option: D
a4 + b4 – a3 – b3 – 2a2b2 + ab
= a4 + b4 – 2a2b2 – a3 – b3 + ab
= (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
= (a – b)2 – a2 + ab – b2 + ab
[∵ a + b = 1]
= (a – b)2 – (a – b)2 = 0
- A complete factorisation of (x4 + 64) is
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x4 + 64 = (x2)2 + (8)2
= (x2 + 8)2 – 2 × 8x2
[∵ a2 + b2 = (a + b)2 – 2ab]
= (x2 + 8)2 – (4x)2
= (x2 + 4x + 8) (x2 – 4x + 8)Correct Option: D
x4 + 64 = (x2)2 + (8)2
= (x2 + 8)2 – 2 × 8x2
[∵ a2 + b2 = (a + b)2 – 2ab]
= (x2 + 8)2 – (4x)2
= (x2 + 4x + 8) (x2 – 4x + 8)
- If k is the largest possible real number such that p4 + q4 = (p2 + kpq + q2)(p2 – kpq + q2), then the value of k is
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p4 + q4 = (p2)2 + (q2)2
= (p2 + q2)2 – 2p2q2
= (p2 + q2)2 – (√2pq)2
= (p2 + q2 + √2pq)(p2 + q2 – √2pq)Correct Option: D
p4 + q4 = (p2)2 + (q2)2
= (p2 + q2)2 – 2p2q2
= (p2 + q2)2 – (√2pq)2
= (p2 + q2 + √2pq)(p2 + q2 – √2pq)
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If a + b + c = 15 and 1 + 1 + 1 = 71 ,then the value of a3 + b3 + c3 – 3abc is a b c abc
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a + b + c = 15,
1 + 1 + 1 = 71 a b c abc ⇒ bc + ac + ab = 71 abc abc
⇒ ab + bc + ca = 71
∴ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
= (a + b + c) {(a + b + c)2 – 3 (ab + bc + ac)}
= 15 (152 – 3 × 71)
= 15 (225 – 213) = 15 × 12
= 180Correct Option: B
a + b + c = 15,
1 + 1 + 1 = 71 a b c abc ⇒ bc + ac + ab = 71 abc abc
⇒ ab + bc + ca = 71
∴ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
= (a + b + c) {(a + b + c)2 – 3 (ab + bc + ac)}
= 15 (152 – 3 × 71)
= 15 (225 – 213) = 15 × 12
= 180
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If 
a + 1 
2 = 3, then the value of a18 + a12 + a6 + 1 is a
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a + 1 2 = 3 a ⇒ a + 1 = √3 a
On cubing both sides,a + 1 3 = (√3)3 a ⇒ a3 + 1 + 3 a + 1 = 3√3 a3 a ⇒ a3 + 1 + 3√3 = 3√3 a3 ⇒ a3 + 1 = 3√3 − 3√3 = 0 a3 ∴ a6 − 1 a6 ⇒ a6 + 1 = 0 a3
⇒ a6 + 1 = 0
∴ a18 + a12 + a6 + 1
= a12 (a6 + 1) + 1 (a6 + 1)
= (a6 + 1)(a12 + 1) = 0Correct Option: C
a + 1 2 = 3 a ⇒ a + 1 = √3 a
On cubing both sides,a + 1 3 = (√3)3 a ⇒ a3 + 1 + 3 a + 1 = 3√3 a3 a ⇒ a3 + 1 + 3√3 = 3√3 a3 ⇒ a3 + 1 = 3√3 − 3√3 = 0 a3 ∴ a6 − 1 a6 ⇒ a6 + 1 = 0 a3
⇒ a6 + 1 = 0
∴ a18 + a12 + a6 + 1
= a12 (a6 + 1) + 1 (a6 + 1)
= (a6 + 1)(a12 + 1) = 0
