Question

(x-2y)cube + (2y-3z)cube + (3z-x)cube​

Answer 1

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${(x - 2y)}^{3} + {(2y - 3z)}^{3} + {(3z - x)}^{3}$

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Here,this identity is used :

$\bold{\boxed{{(a - b)}^{3} = {a}^{3} + {b}^{ 3} - 3ab(a - b)}}$

$\bold{{(x - 2y)}^{3} = {x}^{3} + {(2y)}^{3} - 3(x)(2y)(x - 2y)}$

$= {x}^{3} + 8 {y}^{3} - 6xy(x - 2y)$

$\bold{\red{= {x}^{3} + 8 {y}^{3} - 6 {x}^{2} y + 12x {y}^{2} }}$

$\bold{= > {(2y - 3z)}^{3} = {(2y)}^{3} + {(3z)}^{3} - 3(2y)(3z)(2y - 3z)}$

$\bold{\green{= 8 {y}^{3} + 27 {z}^{3} - 24 {y}^{2} z + 36y {z}^{2}}}$

$\bold{= > {(3z - x)}^{3} = {(3z)}^{3} + {x}^{3} - 3(3z)(x)(3z - x)}$

$= > 27 {z}^{3} + {x}^{3} - 9xz(3z - x)$

$\bold{\blue{= 27 {z}^{3} + {x}^{3} - 27x {z}^{2} + 9 {x}^{2} z}}$

Now adding all three values we obtain:

${(x - 2y)}^{3} + {(2y - 3z)}^{3} + {(3z - x)}^{3} = {\red{{x}^{3} + 8 {y}^{3} - 6 {x}^{2} y + 12x {y}^{2} }}+ {\green{8 {y}^{3} + 27 {z}^{3} - 24 {y}^{2} z + 36y {z}^{2} }}+ {\blue{27 {z}^{3} + {x}^{3} - 27x {z}^{2} + 9 {x}^{2} z}}$

$= 2 {x}^{3} + 16 {y}^{3} + 54 {z}^{3} - 6 {x}^{2} y + 12x {y}^{2} - 24 {y}^{2} z + 36y {z}^{2} - 27x {z}^{2} + 9 {x}^{2} z$

Taking variable common from like terms:

$\bold{= > 2 {x}^{3} + 16 {y}^{3} + 54 {z}^{3} - 6xy(x + 2y) - 6yz(4y - 6z) - 9xz(3z - x)}$

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