(B) We can express $999$ as $(1000 - 1)$.Using the algebraic identity $(x - y)^{3} = x^{3} - y^{3} - 3xy(x - y)$,where $x = 1000$ and $y = 1$:$(1000 -

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(B) We can express $999$ as $(1000 - 1)$.Using the algebraic identity $(x - y)^{3} = x^{3} - y^{3} - 3xy(x - y)$,where $x = 1000$ and $y = 1$:$(1000 - 1)^{3} = (1000)^{3} - (1)^{3} - 3(1000)(1)(1000 - 1)$$= 1,000,000,000 - 1 - 3000(999)$$= 1,000,000,000 - 1 - 2,997,000$$= 1,000,000,000 - 2,997,001$$= 997,002,999$

Class 9 Mathematics Polynomials Textbook - Polynomials

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Evaluate using suitable identities: $(999)^{3}$

A
$997208999$
B
$997002999$
C
$997002900$
D
$977002299$

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Textbook - Polynomials

Verify whether the following are zeroes of the polynomial indicated against them:
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Medium

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Factorise each of the following: $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

Easy

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Expand each of the following,using suitable identities: $(-2x + 5y - 3z)^2$

Medium

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Factorise: $12 x^{2}-7 x+1$

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Give one example each of a binomial of degree $35$ and of a monomial of degree $100$.

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Evaluate using suitable identities: $(999)^{3}$

Similar Questions

Verify whether the following are zeroes of the polynomial indicated against them:$p(x) = (x + 1)(x - 2)$,$x = -1, 2$

Factorise each of the following: $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

Expand each of the following,using suitable identities: $(-2x + 5y - 3z)^2$

Factorise: $12 x^{2}-7 x+1$

Give one example each of a binomial of degree $35$ and of a monomial of degree $100$.

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Verify whether the following are zeroes of the polynomial indicated against them:
$p(x) = (x + 1)(x - 2)$,$x = -1, 2$