Question

Factorise:
$27x {}^{3} + y {}^{3} + {z}^{3} - 9xyz$
Please tell me the answer of this problem, Tomorrow is my Exam
Please..........

Answer 1

Using a^3 + b^3 + c^3 - 3abc  = (a+b+c+)(a^2+b^2+c^2 -ab-bc-ca).

27x^3 +y^3 +z^3 - 9xyz

(3x)^3 + (y)^3 + (z)^3 -  3(3x*y*z)
=>(3x+y+z)(9x^2+y^2+z^2 -3xy-yz-3zx)

Answer 2

✍HERE IS YOUR ANSWER...⤵⤵
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$\underline{SOLUTION}$ ➡

$\underline{We \: \: know\: \: that}$

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

$\underline{Using \: \: this \: \: formula \:, \: we \: \: get,}$

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

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