Which of the following identity can be used to solve 1001 nbsp- 999-a-b-2 - a-2 - 2ab - b-2 nbsp-a - b- - nbsp-a - b- nbsp- a-2 - b-2 nbsp-a-b-2 - a-2 -2ab - b-2 nbsp-x-y-3 nbsp- x-3 - 3x-2y - 3xy-2 - y-3

The correct option is B (a + b) × nbsp;(a b) nbsp;= a^2 b^2 nbsp;1001 = (1000 + 1) 999 = (1000 1) (1000 + 1)(1000 1) = 1000^2 1^2 [Identity (a ...

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Standard IX

Mathematics

Polynomial Expressions

Which of the ...

Question

Which of the following identity can be used to solve 1001 $\times$ 999?

(a + b) \times (a - b) = a^{2}

b^{2}

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{(a + b)}^{2}

a^{2}

2 a b

b^{2}

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{(a - b)}^{2}

a^{2}

2 a b

b^{2}

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(x + y)^{3}

x^{3}

+ 3

x^{2}

y

+ 3

x

y^{2}

y^{3}

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Solution

The correct option is A $(a + b) \times (a - b) = a^{2}$ - $b^{2}$
1001 = (1000 + 1)
999 = (1000 - 1)
(1000 + 1)(1000 - 1) = $1000^{2}$ - $1^{2}$ [Identity (a + b) $\times$ (a - b) = $a^{2}$ - $b^{2}$ is used]
= 1000000 -1 = 999999

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Similar questions

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Q. 2. Solve 5 examples for each identity that you use for factorization
i)a2+b2 +2ab= (a+b)2
ii)a2+b2-2ab=(a-b)2
iii)a3+b3 = (a+b)(a2+b2-ab)

Solve:

(a^{2} + b^{2} +2ab) - (a^{2} + b^{2} - 2ab)

Q. Question 3

Which of the following is correct?
a)

(a - b)^{2} = a^{2} + 2 a b - b^{2}

(a - b)^{2} = a^{2} - 2 a b + b^{2}

(a - b)^{2} = a^{2} - b^{2}

(a + b)^{2} = a^{2} + 2 a b - b^{2}

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Which of the following identity can be used to solve 1001 × 999?

The correct option is A (a+b)×(a−b)=a2 - b2 1001 = (1000 + 1) 999 = (1000 - 1) (1000 + 1)(1000 - 1) = 10002 - 12 [Identity (a + b) × (a - b) = a2 - b2 is used] = 1000000 -1 = 999999

Which of the following identity can be used to solve 1001 $\times$ 999?

The correct option is A $(a + b) \times (a - b) = a^{2}$ - $b^{2}$
1001 = (1000 + 1)
999 = (1000 - 1)
(1000 + 1)(1000 - 1) = $1000^{2}$ - $1^{2}$ [Identity (a + b) $\times$ (a - b) = $a^{2}$ - $b^{2}$ is used]
= 1000000 -1 = 999999