Algebra
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If x = a (b – c), y = b (c – a) and z = c (a – b), then 
x 
3 + x 3 + x 3 = a b c
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x = b - c ; y = b - c ; z = b - c a b c
Again, b – c + c – a + a – b = 0∴ x 3 + x 3 + x 3 = ( b - c )3 + ( c - a )3 + ( c - a )3 a b c
= 3 (b – c) (c – a) (a – b)Required answer = 3xyz abc
Correct Option: C
x = b - c ; y = b - c ; z = b - c a b c
Again, b – c + c – a + a – b = 0∴ x 3 + x 3 + x 3 = ( b - c )3 + ( c - a )3 + ( c - a )3 a b c
= 3 (b – c) (c – a) (a – b)Required answer = 3xyz abc
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If xy (x + y) = 1, then the value of 1 - x3 - y3 is : x3y3
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xy (x + y) = 1
⇒ x + y = 1 xy
On cubing both sides,⇒ x3 + y3 + 3xy (x + y) = 1 x3y3 ⇒ x3 + y3 + 3xy × 1 = 1 xy x3y3 ⇒ 1 - x3 - y3 = 3 x3y3 Correct Option: C
xy (x + y) = 1
⇒ x + y = 1 xy
On cubing both sides,⇒ x3 + y3 + 3xy (x + y) = 1 x3y3 ⇒ x3 + y3 + 3xy × 1 = 1 xy x3y3 ⇒ 1 - x3 - y3 = 3 x3y3
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If x4 + 1 = 119 ,and x > l, then the value of x3 - 1 is x4 x3
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Using Rule1 and 8,
x4 + 1 = 119 x4 ⇒ x2 + 1 2 - 2 = 119 x2 ⇒ x2 + 1 2 = 121 x2 ⇒ x2 + 1 = 11 x2 ⇒ x - 1 2 + 2 = 11 x ⇒ x - 1 2 = 9 x ⇒ x - 1 = 3 x ∴ x3 - 1 - 3 x - 1 = 27 x3 x ⇒ x3 - 1 - 3 × 3 = 27 x3 ⇒ x3 - 1 = 27 + 9 = 36 x3
Correct Option: D
Using Rule1 and 8,
x4 + 1 = 119 x4 ⇒ x2 + 1 2 - 2 = 119 x2 ⇒ x2 + 1 2 = 121 x2 ⇒ x2 + 1 = 11 x2 ⇒ x - 1 2 + 2 = 11 x ⇒ x - 1 2 = 9 x ⇒ x - 1 = 3 x ∴ x3 - 1 - 3 x - 1 = 27 x3 x ⇒ x3 - 1 - 3 × 3 = 27 x3 ⇒ x3 - 1 = 27 + 9 = 36 x3
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If 3x + 1 = 5 , then the value of 8x3 + 1 is 2x 27x3
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Using Rule 8,
3x + 1 = 5 2x
On multiplying both sides by ( 2 / 5 )⇒ 2x + 1 = 10 3x 3
Cubing both sides,∴ 8x3 + 1 + 3 × 2x × 1 × 2x + 1 = 1000 27x3 3x 3x 27 ⇒ 8x3 + 1 + 3 × 3 = 27 8x3 ⇒ 8x3 + 1 + 2 × 10 = 1000 27x3 3 27 ⇒ 8x3 + 1 = 1000 - 20 27x3 27 3 ⇒ 8x3 + 1 = 1000 - 180 = 820 = 30 10 27x3 27 27 27 Correct Option: B
Using Rule 8,
3x + 1 = 5 2x
On multiplying both sides by ( 2 / 5 )⇒ 2x + 1 = 10 3x 3
Cubing both sides,∴ 8x3 + 1 + 3 × 2x × 1 × 2x + 1 = 1000 27x3 3x 3x 27 ⇒ 8x3 + 1 + 3 × 3 = 27 8x3 ⇒ 8x3 + 1 + 2 × 10 = 1000 27x3 3 27 ⇒ 8x3 + 1 = 1000 - 20 27x3 27 3 ⇒ 8x3 + 1 = 1000 - 180 = 820 = 30 10 27x3 27 27 27
- If x + y = z, then the expression x3 + y3 - z3 + 3xyz will be equal to :
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Using Rule 20,
x + y = z ⇒ x + y + (–z) = 0
∴ x3 + y3 - z3 + 3xyz = x3 + y3 + ( - z )3 – 3x.y (–z) = 0Correct Option: A
Using Rule 20,
x + y = z ⇒ x + y + (–z) = 0
∴ x3 + y3 - z3 + 3xyz = x3 + y3 + ( - z )3 – 3x.y (–z) = 0
