Question

factorise 2√2a^3+8b^3 -27c^3+18√2abc

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\red { 2\sqrt{2} a^{3} + 8b^{3} - 27c^{3} + 18\sqrt{2}abc}$

$\green {=(\sqrt{2}a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt{2}ab+6bc+3\sqrt{2}ca)}$

Step-by-step explanation:

$2\sqrt{2} a^{3} + 8b^{3} - 27c^{3} + 18\sqrt{2}abc$

$=(\sqrt{2}a)^{3}+(2b)^{3}+(-3c)^{3}-3\times (\sqrt{2}a)(2b)(-3c)}$

$\pink { x^{3}+y^{3}+z^{3}-3xyz}$

$\orange = { (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)}$

$= (\sqrt{2}a+2b-3c)[(\sqrt{2}a)^{2}+(2b)^{2}+(-3c)^{2} - (\sqrt{2}a)(2b)-(2b)(-3c)-(-3c)(\sqrt{2}a)]$

$= (\sqrt{2}a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt{2}ab+6bc+3\sqrt{2}ca)$

Therefore.,

$\red { 2\sqrt{2} a^{3} + 8b^{3} - 27c^{3} + 18\sqrt{2}abc}$

$\green {=(\sqrt{2}a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt{2}ab+6bc+3\sqrt{2}ca)}$

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