Factorise : 8 x^{3}+y^{3}+27 z^{3}-18 x y z | vedclass

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(N/A) We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.Here,$a = 2x$,$b = y$,and $c = 3z$.Substituting th

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(N/A) We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Here,$a = 2x$,$b = y$,and $c = 3z$.
Substituting these values into the identity:
$8x^{3} + y^{3} + 27z^{3} - 18xyz = (2x)^{3} + (y)^{3} + (3z)^{3} - 3(2x)(y)(3z)$.
Applying the formula:
$= (2x + y + 3z)((2x)^{2} + (y)^{2} + (3z)^{2} - (2x)(y) - (y)(3z) - (2x)(3z))$.
Simplifying the terms inside the second bracket:
$= (2x + y + 3z)(4x^{2} + y^{2} + 9z^{2} - 2xy - 3yz - 6xz)$.

Factorise $: 8 x^{3}+y^{3}+27 z^{3}-18 x y z$

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