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`(a + b)^2 = a^2 +
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Jan 17, 2022
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8th-maths
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yogita
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chapter9
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`(a – b)^2 = a^2 –
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`p^2q + q^2r + r^2q` is a
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Factorisation of `– 3a^2 + 3ab + 3ac` is `3a (–a – b –
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abc + bca + cab is a
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`(9x – 51) -: 9` is x –
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The value of `(a + 1) (a – 1) (a^2 + 1)` is `a^4 –
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Subtract : `3t^4 – 4t^3 + 2t^2 – 6t + 6` from `– 4t^4 + 8t^3 – 4t^2 – 2t +
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Jan 17, 2022
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Multiply the following: `– 7pq^2r^3, –
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Multiply the following: `15xy^2,
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Expand the following, using suitable identities. `(xy +
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Expand the following, using suitable identities. `(7x +
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Jan 17, 2022
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Using suitable identities, evaluate the following.
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Factorise the following, using the identity `a^2 + 2ab + b^2 = (a + b)^2` `a^2x^2 + 2ax +
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Jan 17, 2022
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marks3
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Factorise the following, using the identity `a^2 + 2ab + b^2 = (a + b)^2` `a^2x^2 + 2abx +
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Factorise the following, using the identity `a^2 – 2ab + b^2 = (a – b)^2`. `9x^2 – 12x +
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Jan 17, 2022
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sumit
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Factorise the following. `x^2 + 15x +
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Jan 17, 2022
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Factorise the following. `x^2 = 17x + 60`
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Jan 17, 2022
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Factorise the following. `p^2 – 13p –
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Jan 17, 2022
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sumit
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Factorise the following using the identity `a^2 – b^2 = (a + b) (a – b)`. `y^4 –
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Jan 17, 2022
in
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nikhil
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