Question

Factorise: 27x 3 +y 3 +z 3 -9 xyz.

Answer 1

Solution:

Given 27x³+y³+z³-9xyz

= (3x)³+y³+z³-3(3x)yz

/*By algebraic identity:

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

= (3x+y+z)[(3x)²+y²+z²-(3x)y-yz-z(3x)]

=(3x+y+z)(9x²+y²+z²-3xy-yz-3zx)

Therefore,

$27x^{3}+y^{3}+z^{3}-9xyz\\=(3x+y+z)(9x^{2}+y^{2}+z^{2}-3xy-yz-3zx)$

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Answer 2

HERE IS YOUR ANSWER..⬇⬇


$27x {}^{3} + y {}^{3} + z {}^{3} - 9xyz \\ \\ = (3x) {}^{3} + (y) {}^{3} + (z) {}^{3} - 3 \times 3x \times y \times z \\ \\ = (3x + y+ z)((3x) {}^{2} + y {}^{2} + z {}^{2} - 3x \times y - y \times z - z \times 3x) \\ \\ = (3x + y + z)(9x {}^{2} + y {}^{2} + z {}^{2} - 3xy - yz - 3zx)$


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