Algebra
- If a + b + c = 6 and ab + bc + ca = 11, then the value of bc (b + c) + ca (c + a) + ab (a + b) + 3abc is
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a + b + c = 6 and ab + bc + ca = 11
∴ bc (b + c) + ca (c + a) + ab (a + b) + 3abc
= bc (b + c) + abc + ca (c + a) + abc + ab (a + b) + abc
= bc (a + b + c) + ca (a + b + c) + ab (a + b + c)
= (a + b + c) (bc + ca + ab)
= 6 × 11 = 66Correct Option: B
a + b + c = 6 and ab + bc + ca = 11
∴ bc (b + c) + ca (c + a) + ab (a + b) + 3abc
= bc (b + c) + abc + ca (c + a) + abc + ab (a + b) + abc
= bc (a + b + c) + ca (a + b + c) + ab (a + b + c)
= (a + b + c) (bc + ca + ab)
= 6 × 11 = 66
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If x = 3 , then the value of 27x3 – 54x2 + 36x – 11 is 2
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x = 3 (Given) 2
∴ 27x3 – 54x2 + 36x – 11
= (3x)3 – 3 × (3x)2 × 2 + 3 × 3x
(2)2 – (2)3 – 3
= (3x – 2)3 – 3
[∵ (a – b)3 = a3 – 3a2 b + 3ab2 – b3]= 
3 × 3 − 2 
3 − 3 2 = 9 − 2 3 − 3 2 = 9 − 4 3 − 3 2 = 5 3 − 3 2 = 125 − 3 8 = 125 − 24 8 = 101 8 12 5 8 Correct Option: D
x = 3 (Given) 2
∴ 27x3 – 54x2 + 36x – 11
= (3x)3 – 3 × (3x)2 × 2 + 3 × 3x
(2)2 – (2)3 – 3
= (3x – 2)3 – 3
[∵ (a – b)3 = a3 – 3a2 b + 3ab2 – b3]= 3 × 3 − 2 3 − 3 2 = 9 − 2 3 − 3 2 = 9 − 4 3 − 3 2 = 5 3 − 3 2 = 125 − 3 8 = 125 − 24 8 = 101 8 12 5 8
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If x + 1 2 = 3, then the value of x3 + 1 is x x3
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x + 1 2 = 3 x ∴ x + 1 = √3 x
On cubing both sides,x + 1 3 = (√3)3 x ⇒ x3 + 1 + 3 x + 1 = 3√3 x3 x ⇒ x3 + 1 + 3√3 = 3√3 x3 ⇒ x3 + 1 = 3√3 − 3√3 = 0 x3 Correct Option: A
x + 1 2 = 3 x ∴ x + 1 = √3 x
On cubing both sides,x + 1 3 = (√3)3 x ⇒ x3 + 1 + 3 x + 1 = 3√3 x3 x ⇒ x3 + 1 + 3√3 = 3√3 x3 ⇒ x3 + 1 = 3√3 − 3√3 = 0 x3
- If l + m + n = 9 and l2 + m2 + n2 31, then the value of (lm + mn + nl) will be
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l2 + m2 + n2 31
l + m + n = 9
On squaring both sides,
(l + m + n)2 = 81
⇒ l2 + m2 + n2 + 2 (lm + mn + nl) = 81
⇒ 31 + 2(lm + mn + nl) = 81
⇒ 2(lm + mn + nl) = 81 – 31 = 50⇒ lm + mn + nl = 50 = 25 2 Correct Option: C
l2 + m2 + n2 31
l + m + n = 9
On squaring both sides,
(l + m + n)2 = 81
⇒ l2 + m2 + n2 + 2 (lm + mn + nl) = 81
⇒ 31 + 2(lm + mn + nl) = 81
⇒ 2(lm + mn + nl) = 81 – 31 = 50⇒ lm + mn + nl = 50 = 25 2
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If a + b = 1, the value of a3 + b3 is equal to b a
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a + b = 1 ⇒ a2 + b2 = 1 b a ab
⇒ a2 + b2 = ab
⇒ a2 – ab + b2
= 0
∴ a3 + b3 = (a + b) (a2 – ab + b2) = 0Correct Option: A
a + b = 1 ⇒ a2 + b2 = 1 b a ab
⇒ a2 + b2 = ab
⇒ a2 – ab + b2
= 0
∴ a3 + b3 = (a + b) (a2 – ab + b2) = 0
