Question

factorise 27x^3+y^3+z^3-9xyz

Answer 1

Hello dear friend .
your answer is here
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Sol.

$given \: \: 27 {x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ = > (3 {x)}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ using \: \: identity \: - \\ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z) \\ ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ \\ = > (3 {x})^{3} + {y}^{3} + {z}^{3} - 3(3x)(y)(z) \\ \\ = > (3x + y + z) = (3 {x})^{2} + {y}^{2} + {z}^{2} - \\ (3x)(y) - (y)(z) - (z)(3x) \\ \\ = > (3x + y + z)(9 {x}^{2} + {y}^{2} + {z}^{2} - 3xy - yz - 3xz) \\ \\ answer$
Hope it's helps you.
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Answer 2

Hey!

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Question = 27x³ + y³ + z³ - 9xyz
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Identity :-

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x³ + y³ + z³ - 3xyz = (x + y + z) × (x² + y² + z² - xy - yz - zx)

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= 27x³ + y³ + z³ - 9xyz

= (3x)³ + y³ + z³ - (3x) × y × z

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•Now use the given Identity :-

=> (3x + y + z) = (3x)² + y² + z² - (3x) × y - (y) × (z) - (z) × (3x)

=> (3x + y + z) × (9x² + y² + z² - 3xy - yz - 3xz)

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Regards :)

Cybary

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