Math, asked by ronaldochristanio216, 1 month ago

2(2)^1/2a^3+8b^3-27c^3+18(2)^1/2abc
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Answers

Answered by soumyavaishnavivinu
0

Answer:

2

2

a

3

+8b

3

−27c

3

+18

2

abc

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

2

a+2b−3c)(2a

2

+4b

2

+9c

2

−2

2

ab+6bc+3

2

ca)

Step-by-step explanation:

2\sqrt{2} a^{3} + 8b^{3} - 27c^{3} + 18\sqrt{2}abc2

2

a

3

+8b

3

−27c

3

+18

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}x

3

+y

3

+z

3

−3xyz

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

2

a+2b−3c)[(

2

a)

2

+(2b)

2

+(−3c)

2

−(

2

a)(2b)−(2b)(−3c)−(−3c)(

2

a)]

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

2

a+2b−3c)(2a

2

+4b

2

+9c

2

−2

2

ab+6bc+3

2

ca)

Therefore.,

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

2

a

3

+8b

3

−27c

3

+18

2

abc

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

2

a+2b−3c)(2a

2

+4b

2

+9c

2

−2

2

ab+6bc+3

2

ca)

Step-by-step explanation:

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