Algebra
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If p2 + q2 = 1 , then the value of (p6 + q6) is q2 p2
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p2 + q2 = 1 q2 p2 ⇒ p4 + q4 = 1 p2q2
⇒ p4 + q4 = p2q2
⇒ p4 + q4 - p2q2 = 0 ...... (i)
∴ p6 + q6 = (p2)3 + (q2)3
p6 + q6 = (p2 + q2)(p4 + q4 - p2q2)
[ ∴ a3 + b3 = (a + b)(a2 + b2 - ab) ]
p6 + q6 = (p2 + q2) × 0 = 0Correct Option: A
p2 + q2 = 1 q2 p2 ⇒ p4 + q4 = 1 p2q2
⇒ p4 + q4 = p2q2
⇒ p4 + q4 - p2q2 = 0 ...... (i)
∴ p6 + q6 = (p2)3 + (q2)3
p6 + q6 = (p2 + q2)(p4 + q4 - p2q2)
[ ∴ a3 + b3 = (a + b)(a2 + b2 - ab) ]
p6 + q6 = (p2 + q2) × 0 = 0
- If a + b – c = 0 then the value of 2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4
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Expression = 2b2c2 + 2c2a2 + 2a2b2 - a4 - b4 - c4
Expression = 4b2c2 - (2b2c2 - 2c2a2 - 2a2b2 + a4 + b4 + c4)
Expression = (2bc)2 - (a2 - b2 - c2)2
Expression = (2bc + a2 - b2 - c2)(2bc - a2 + b2 + c2)
Expression = {a2 - (b2 + c2 - 2bc){(b2 + c2 + 2bc) - a2}
Expression = { a2 - (b - c)2 }{ (b + c)2 - a2 }
Expression = (a – b + c) (a + b – c)(a + b + c) (b + c – a)
If a + b – c = 0,
∴ Expression = 0Correct Option: B
Expression = 2b2c2 + 2c2a2 + 2a2b2 - a4 - b4 - c4
Expression = 4b2c2 - (2b2c2 - 2c2a2 - 2a2b2 + a4 + b4 + c4)
Expression = (2bc)2 - (a2 - b2 - c2)2
Expression = (2bc + a2 - b2 - c2)(2bc - a2 + b2 + c2)
Expression = {a2 - (b2 + c2 - 2bc){(b2 + c2 + 2bc) - a2}
Expression = { a2 - (b - c)2 }{ (b + c)2 - a2 }
Expression = (a – b + c) (a + b – c)(a + b + c) (b + c – a)
If a + b – c = 0,
∴ Expression = 0
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If 4a - 4 + 3 = 0 then the value of : a3 - 1 + 3 = ? a a3
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4a - 4 = -3 a
On dividing by 4,⇒ a - 1 = -3 a 4 ∴ a3 - 1 = a - 1 3 + 3 . a . 1 a - 1 a3 a a a ⇒ a3 - 1 = -3 3 + 3 × -3 a3 4 4 ⇒ a3 - 1 = - 27 - 9 a3 64 4 ⇒ a3 - 1 = -27 - 144 = -171 a3 64 64 ∴ a3 - 1 + 3 = -171 + 3 = -171 + 192 a3 64 64 a3 - 1 + 3 = 21 a3 64
Correct Option: C
4a - 4 = -3 a
On dividing by 4,⇒ a - 1 = -3 a 4 ∴ a3 - 1 = a - 1 3 + 3 . a . 1 a - 1 a3 a a a ⇒ a3 - 1 = -3 3 + 3 × -3 a3 4 4 ⇒ a3 - 1 = - 27 - 9 a3 64 4 ⇒ a3 - 1 = -27 - 144 = -171 a3 64 64 ∴ a3 - 1 + 3 = -171 + 3 = -171 + 192 a3 64 64 a3 - 1 + 3 = 21 a3 64
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If x2 + x = 5 then the value of (x + 3)3 + 1 is (x + 3)3
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x2 + x = 5 (Given)
Let, x + 3 = a
∴ 1 = 1 x + 3 a Now , a + 1 = (x + 3) + 1 a (x + 3) a + 1 = (x + 3)2 + 1 a (x + 3) a + 1 = x2 + 6x + 9 + 1 a x + 3 a + 1 = x2 + x + 5x + 10 a x + 3 a + 1 = 5 + 5x + 10 a x + 3 a + 1 = 5x + 15 = 5(x + 3) = 5 a x + 3 x + 3 ∴ a3 + 1 = a + 1 3 - 3 . a . 1 a + 1 a3 a a a
Required answer = (5)3 – 3 × 5 = 125 – 15 = 110
Correct Option: B
x2 + x = 5 (Given)
Let, x + 3 = a
∴ 1 = 1 x + 3 a Now , a + 1 = (x + 3) + 1 a (x + 3) a + 1 = (x + 3)2 + 1 a (x + 3) a + 1 = x2 + 6x + 9 + 1 a x + 3 a + 1 = x2 + x + 5x + 10 a x + 3 a + 1 = 5 + 5x + 10 a x + 3 a + 1 = 5x + 15 = 5(x + 3) = 5 a x + 3 x + 3 ∴ a3 + 1 = a + 1 3 - 3 . a . 1 a + 1 a3 a a a
Required answer = (5)3 – 3 × 5 = 125 – 15 = 110
- If x2 + y2 + z2 = 2 (x + z – 1), then the value of :
x3 + y3 + z3 = ?
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x2 + y2 + z2 = 2(x + z – 1)
⇒ x2 + y2 + z2 = 2x + 2z – 2
⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
⇒ (x - 1)2 + y2 + (z - 1)2 = 0
∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
∴ x – 1 = 0 ⇒ x = 1
y = 0
z – 1 = 0 ⇒ z = 1
∴ x3 + y3 + z3 = 1 + 0 + 1 = 2Correct Option: A
x2 + y2 + z2 = 2(x + z – 1)
⇒ x2 + y2 + z2 = 2x + 2z – 2
⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
⇒ (x - 1)2 + y2 + (z - 1)2 = 0
∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
∴ x – 1 = 0 ⇒ x = 1
y = 0
z – 1 = 0 ⇒ z = 1
∴ x3 + y3 + z3 = 1 + 0 + 1 = 2