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Factorize \[(a+2b-3c)^3 +(b+2c-3a)^3+(c+2a-3b)^3\]

Note by Abhishek Alva 1 year, 9 months ago

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Let \((a+2b-3c)=x,(b+2c-3a)=y\) and \((c+2a-3b)=z.\)We can clearly see that \(x+y=z=0.\) So,\(x^3+y^3+z^3=3(a+2b-3c)(b+2c-3a)(c+2a-3b).\)

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ya your are right

3(a+2b-3c)(b+2c-3a)(c+2a-3b)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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`[example link](https://brilliant.org)`

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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TopNewestLet \((a+2b-3c)=x,(b+2c-3a)=y\) and \((c+2a-3b)=z.\)We can clearly see that \(x+y=z=0.\)

So,\(x^3+y^3+z^3=3(a+2b-3c)(b+2c-3a)(c+2a-3b).\)

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ya your are right

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3(a+2b-3c)(b+2c-3a)(c+2a-3b)

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