Question

Factorise: (x-3y)^3+ (3y-7z)^3 + (7z - x)^3​

Answer 1

$if \: we \: assume \: the \: three \: terms \: to \: be \: a \: b \: and \: c \: \\ their \: sum \: will \: be \: equal \: to \: zero \: and \: in \: that \: \: case \\ {a}^{3} + {b}^{3} + {c}^{3} \: = 3abc \\ \\ = 3(x - 3y)(3y - 7z)(7z - x) \\ = 3(3xy - 7zx - 9 {y}^{2} + 21zy)(7z - x) \\ = 3(21xyz - 49 {z}^{2} x - 63 {y}^{2} z + 147 {z}^{2} y - 3 {x}^{2} y + 7z {x}^{2} + 9 {y}^{2} x - 21xyz)$

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