Question

Factorise:8x³+y³+27z³-18xyz.

Answer 1

Answer:

8x^3+y^3+27z^3-18xyz=

((2x)^3+(y^3)+(3z^3))((2x)^2+y^2+3z^2-(2x)(y)-(y)(3z)-(2x)(3z)

=(2x+y+z)(4x^2+y^2+9z^2-2xy-3yz-6zx

hope it helps

Step-by-step explanation:

Answer 2

$\red{ 8x^{3}+y^{3}+27z^{3}-18xyz$

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

$\pink { a^{3}+b^{3}+c^{3}-3abc }\\\pink {= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca )}$

$Here , a = 2x ,\:b = y \:and \: c = 3z$

$= (2x + y + 3z)[(2x)^{2}+y^{2}+(3z)^{2}-2x\times y - y \times 3z - 3z \times 2x ] \\= (2x+y+3z)(4x^{2}+y^{2}+9z^{2}-2xy-3yz-6xz)$

Therefore.,

$\red{ Factors \:of \:8x^{3}+y^{3}+27z^{3}-18xyz}\\\green {= (2x+y+3z)(4x^{2}+y^{2}+9z^{2}-2xy-3yz-6xz) }$

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