Algebra
- x2 + y2 + z2 = 14 and xy + yz + zx = 11, then the value of (x + y + z)2 is
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Given,
x2 + y2 + z2 = 14
xy + yz + zx = 11
∴ (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
= 14 + 2 × 11
= 14 + 22 = 36Correct Option: C
Given,
x2 + y2 + z2 = 14
xy + yz + zx = 11
∴ (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
= 14 + 2 × 11
= 14 + 22 = 36
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If x3 + 1 = 110, then find the value of x + 1 . x3 x
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a3 + b3 = (a + b)3 – 3ab (a + b)
∴ x3 + 1 = 110 x3 ⇒ 
x + 1 
3 − 3 x + 1 x x
= (5)3 – 3 × 5⇒ x + 1 = 5 x Correct Option: D
a3 + b3 = (a + b)3 – 3ab (a + b)
∴ x3 + 1 = 110 x3 ⇒ x + 1 3 − 3 x + 1 x x
= (5)3 – 3 × 5⇒ x + 1 = 5 x
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If x4 + 1 = 119, then the value of x − 1 is x4 x
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x4 + 1 = 119 x4 ⇒ x2 + 1 2 – 2 = 119 x2 ⇒ x2 + 1 2 = 119 + 2 = 121 x2 ⇒ x2 + 1 2 = 112 x2 ⇒ x2 + 1 = 11 x2 ∴ x − 1 2 + 2 = 11 x ⇒ x − 1 2 = 11 – 2 = 9 = 32 x ⇒ x − 1 = 3 x Correct Option: D
x4 + 1 = 119 x4 ⇒ x2 + 1 2 – 2 = 119 x2 ⇒ x2 + 1 2 = 119 + 2 = 121 x2 ⇒ x2 + 1 2 = 112 x2 ⇒ x2 + 1 = 11 x2 ∴ x − 1 2 + 2 = 11 x ⇒ x − 1 2 = 11 – 2 = 9 = 32 x ⇒ x − 1 = 3 x
- If m + n = 1, then the value of m3 + n3 + 3mn is equal to
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m3 + n3 + 3mn
= m3 + n3 + 3mn(m + n) [∵ m + n =1]
= (m + n)3 = 1Correct Option: B
m3 + n3 + 3mn
= m3 + n3 + 3mn(m + n) [∵ m + n =1]
= (m + n)3 = 1
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If a + 1 2 = 3, then the value of a6 − 1 will be a a6
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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 = a3 + 1 a3 − 1 = 0 a3 a3 Correct 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 = a3 + 1 a3 − 1 = 0 a3 a3
