Algebra
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If x + 1 = 2 ,find the value of 8x3 + 1 . 2x x3
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Using Rule 8,
x + 1 = 2 2x ⇒ 2x + 2 = 2 × 2 = 4 2x ⇒ 2x + 1 = 4 x
On cubing both sides,∴ 8x3 + 1 + 3 × 2x × 1 
2x + 1 
= 64 x3 x x ⇒ 8x3 + 1 + 6 × 4 = 64 x3 ⇒ 8x3 + 1 = 64 – 24 = 40 x3 Correct Option: C
Using Rule 8,
x + 1 = 2 2x ⇒ 2x + 2 = 2 × 2 = 4 2x ⇒ 2x + 1 = 4 x
On cubing both sides,∴ 8x3 + 1 + 3 × 2x × 1 2x + 1 = 64 x3 x x ⇒ 8x3 + 1 + 6 × 4 = 64 x3 ⇒ 8x3 + 1 = 64 – 24 = 40 x3
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If x = a + 1 and y = a - 1 then the value of x4 + y4 - 2x2y2 is a a
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Given , a + 1 + 1 = 0 a
x4 + y4 - 2x2y2 = ( x2 - y2 )2
= [ ( x + y )( x - y ) ]2= 
a + 1 + a - 1 a + 1 - a + 1 
2 a a a a = 2a × 2 2 = 16 a
Correct Option: C
Given , a + 1 + 1 = 0 a
x4 + y4 - 2x2y2 = ( x2 - y2 )2
= [ ( x + y )( x - y ) ]2= a + 1 + a - 1 a + 1 - a + 1 2 a a a a = 2a × 2 2 = 16 a
- If x1 / 3 + y1 / 3 = z1 / 3 , then {(x + y – z)3 + 27xyz} equals :
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Using Rule 8,
x1 / 3 + y1 / 3 = z1 / 3 ......(i)
Cubing both sides,
( x1 / 3 + y1 / 3 = z1 / 3 )3 = z
⇒ x + y + 3 x1 / 3 . y1 / 3 ( 1 / 3 + y1 / 3 ) = z
[∴ (a + b)3 = a3 + b3 + 3ab (a + b)]
⇒ x + y – z = - 3 x1 / 3 . y1 / 3 .z1 / 3 ......(ii) [From equation (i)]
∴ (x + y – z)3 + 27xyz = [ - 3x1 / 3 . y1 / 3 . z1 / 3 ]3 + 27xyz [From equation (ii)]
= – 27xyz + 27xyz = 0Correct Option: C
Using Rule 8,
x1 / 3 + y1 / 3 = z1 / 3 ......(i)
Cubing both sides,
( x1 / 3 + y1 / 3 = z1 / 3 )3 = z
⇒ x + y + 3 x1 / 3 . y1 / 3 ( 1 / 3 + y1 / 3 ) = z
[∴ (a + b)3 = a3 + b3 + 3ab (a + b)]
⇒ x + y – z = - 3 x1 / 3 . y1 / 3 .z1 / 3 ......(ii) [From equation (i)]
∴ (x + y – z)3 + 27xyz = [ - 3x1 / 3 . y1 / 3 . z1 / 3 ]3 + 27xyz [From equation (ii)]
= – 27xyz + 27xyz = 0
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If 2p + 1 = 4 , then the value of p3 + 1 is p 8p3
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Using Rule 8,
2p + 1 = 4 p ⇒ p + 1 = 2 2p ∴ p + 1 3 = p3 + 1 + 3 × p × 1 p + 1 2p 8p3 2p 2p ⇒ 8 = p3 + 1 + 3 × 2 8p3 2 ⇒ p3 + 1 = 8 - 3 = 5 8p3 Correct Option: B
Using Rule 8,
2p + 1 = 4 p ⇒ p + 1 = 2 2p ∴ p + 1 3 = p3 + 1 + 3 × p × 1 p + 1 2p 8p3 2p 2p ⇒ 8 = p3 + 1 + 3 × 2 8p3 2 ⇒ p3 + 1 = 8 - 3 = 5 8p3
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If 4b2 + 1 = 2 , then the value of 8b3 + 1 is b2 b3
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Using Rule 1,
2b + 1 2 = 4b2 + 1 + 2 × 2b × 1 = 2 + 4 = 6 b b 2 b ⇒ 2b + 1 = √6 b ∴ 8b3 + 1 = 2b + 1 3 - 3 × 2b × 1 2b + 1 b3 b b b
= ( √6 )3 - 6( √6 ) = 6 √6 - 6 √6 = 0Correct Option: A
Using Rule 1,
2b + 1 2 = 4b2 + 1 + 2 × 2b × 1 = 2 + 4 = 6 b b 2 b ⇒ 2b + 1 = √6 b ∴ 8b3 + 1 = 2b + 1 3 - 3 × 2b × 1 2b + 1 b3 b b b
= ( √6 )3 - 6( √6 ) = 6 √6 - 6 √6 = 0
