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Factorize : (x-2y)^3+(2y-3z)^3+(3z-x)^3...

Factorize : `(x-2y)^3+(2y-3z)^3+(3z-x)^3`

Text Solution

Verified by Experts

Here to factorise `(x−2y)^3+(2y−3z)^3+(3z−x)^3`, we need to assume an another equation
Therefore,
`a^3+b^3+c^3−3abc=(a+b+c)(a^2+b^2+c^2−ab+bc+ca)`.....(1)
And in the above equation is`a+b+c=0`, then`a^3+b^3+c^3=3abc` ....(2)
So, for the above question if we assume`a=(x−2y), b=(2y−3z), c=(3z−x)`
Then, `(x−2y)+(2y−3z)+(3z−x)=0` `[∵a+b+c=0]`
Therefore,
`(x−2y)^3+(2y−3z)^3+(3z−x)^3=3(x−2y)(2y−3z)(3z−x)`
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