Algebra
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If a + b = 1, then the value of a³ + b³ – 2 is b a
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a + b = 1 b a ⇒ a² + b² = 1 ab
⇒ a² + b² = ab
⇒ a² – ab + b² = 0
∴ a³ + b³ – 2 = (a + b) (a² – ab + b²) – 2
= – 2Correct Option: B
a + b = 1 b a ⇒ a² + b² = 1 ab
⇒ a² + b² = ab
⇒ a² – ab + b² = 0
∴ a³ + b³ – 2 = (a + b) (a² – ab + b²) – 2
= – 2
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If x + 1 = √3, then the value of 
x³ + 1 
is : x x³
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x + 1 = √3 x
On cubing both sides,x + 1 1 ³ =( √3)³ x ⇒ x³ + 1 + 3 x + 1 = 3√3 x x ⇒ x³ + 1 + 3√3= 3√3 x³ ⇒ x³ + 1 = 3√3= 3√3 = 0 x³ Correct Option: C
x + 1 = √3 x
On cubing both sides,x + 1 1 ³ =( √3)³ x ⇒ x³ + 1 + 3 x + 1 = 3√3 x x ⇒ x³ + 1 + 3√3= 3√3 x³ ⇒ x³ + 1 = 3√3= 3√3 = 0 x³
- If a + b = 3, then the value of a³ + b³ + 9ab is :
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It is given, a + b = 3
∴ a³ + b³ + 9ab
= a³ + b³ + 3ab × 3
= a³ + b³ + 3ab (a + b)
= (a + b)³ = (3)³ = 27Correct Option: A
It is given, a + b = 3
∴ a³ + b³ + 9ab
= a³ + b³ + 3ab × 3
= a³ + b³ + 3ab (a + b)
= (a + b)³ = (3)³ = 27
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If 6x² – 12x + 1 = 0, then the value of 27x³ + 1 is 8x²
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6x² – 12x + 1 = 0
Þ 6x² + 1 = 12x⇒ 6x² + 1 = 12x 2x 2x ⇒ 3x + 1 = 6 2x
On cubing both sides,⇒ 3x + 1 ³ = (6)³ 2x ⇒ (3x)³ + 1 ³ = 3 × 3x × 1 3x + 1 = 216 2x 2x 2x ⇒ 27x³ + 1 + 9 × 6 = 216 8x³ 2 ⇒ 27x³ + 1 = 216 - 27 = 189 8x³
Correct Option: B
6x² – 12x + 1 = 0
Þ 6x² + 1 = 12x⇒ 6x² + 1 = 12x 2x 2x ⇒ 3x + 1 = 6 2x
On cubing both sides,⇒ 3x + 1 ³ = (6)³ 2x ⇒ (3x)³ + 1 ³ = 3 × 3x × 1 3x + 1 = 216 2x 2x 2x ⇒ 27x³ + 1 + 9 × 6 = 216 8x³ 2 ⇒ 27x³ + 1 = 216 - 27 = 189 8x³
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If x² + 1 = 98(x > 0), then the value of x³ + 1 is 2 x³
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x² + 1 = 98 x² ⇒ x + 1 ² - 2 = 98 x ⇒ x + 1 ² = 98 + 2 = 100 x ⇒ x + 1 = √100 = 10.......(i) x
On cubing both sides,x + 1 ³ = (10)³ = 1000 x ⇒ x³ + 1 + 3 x + 1 = 1000 x³ x ⇒ x³ + 1 + 3 × 10 = 1000 x³ ⇒ x³ + 1 = 1000 - 30 = 970 x³ Correct Option: A
x² + 1 = 98 x² ⇒ x + 1 ² - 2 = 98 x ⇒ x + 1 ² = 98 + 2 = 100 x ⇒ x + 1 = √100 = 10.......(i) x
On cubing both sides,x + 1 ³ = (10)³ = 1000 x ⇒ x³ + 1 + 3 x + 1 = 1000 x³ x ⇒ x³ + 1 + 3 × 10 = 1000 x³ ⇒ x³ + 1 = 1000 - 30 = 970 x³
