Question

factorise (2p-q)^3+(q-3r)^3+(3r-2p)^3 ​

Answer 1

$We \: know \:that ,$

$\pink { If \: x+y+z = 0 , then } \\\blue { x^{3} + y^{3} + z^{3} = 3xyz }$

$Here, x = 2p - q, y = q - 3r \:and \: z = 3r-2p$

$x + y + z \\= 2p-q + q - 3r + 3r - 2p \\= 0$

$Now , x^{3}+y^{3}+z^{3} \\= (2p-q)^{3}+(q-3r)^{3}+(3r-2p)^{3} \\= 3(2p-q)(q-3r)(3r-2p)$

Therefore.,

$\red{ Factors \:of \:(2p-q)^{3}+(q-3r)^{3}+(3r-2p)^{3}} \\\green { = 3(2p-q)(q-3r)(3r-2p)}$

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