Question

please answer this question ​

Answer 1

Step-by-step explanation:

2√2a³+8b³-27c³+18√2abc

=>(√2a)³+(2b)³+(-3c)³-3(√2a)(2b)(-3c)

=>(√2a+2b-3c)(2a²+4b²+9c²-2√2ab+6bc+3√2ca)

And There You Go..!

Answer 2

$\large\underline{\bold{Given \:Question - }}$

Factorise :-

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

$\large\underline{\sf{Solution-}}$

Given expression is

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

Now,

Consider,

$\rm :\longmapsto\: {2 \sqrt{2} a}^{3}$

$\rm \: = \: \: \sqrt{2} \times \sqrt{2} \times \sqrt{2} \times {a}^{3}$

$\rm \: = \: \: {( \sqrt{2} a)}^{3}$

So,

$\rm :\implies\:2 \sqrt{2} {a}^{3} = \: \: {( \sqrt{2} a)}^{3}$

Now, Consider,

$\rm :\longmapsto\: {8b}^{3}$

$\rm \: = \: \:2 \times 2 \times 2 \times {b}^{3}$

$\rm \: = \: \: {(2b)}^{3}$

So,

$\rm :\implies\: {8b}^{3} = \: \: {(2b)}^{3}$

Now, Consider

$\rm :\longmapsto\: {27c}^{3}$

$\rm \: = \: \:3 \times 3 \times 3 \times {c}^{3}$

$\rm \: = \: \: {(3c)}^{3}$

So,

$\rm :\implies\: {27c}^{3} = \: \: {(3c)}^{3}$

So, given expression

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

can be rewritten as

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

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

We know that,

$\boxed{ \sf{ \: {x}^{3}+{y}^{3}+{z}^{3} - 3xyz = (x + y + z)( {x}^{2}+{y}^{2}+{z}^{2} - xy - yz - zx)}}$

So,

Here,

$\rm :\longmapsto\:x = \sqrt{2}a$

$\rm :\longmapsto\:y = 2b$

$\rm :\longmapsto\:z = - 3c$

So, on substituting the values, we get

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

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

Additional Information :-

More Identities to know

$\boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}$

$\boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}$

$\boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\boxed{ \sf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \sf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \sf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}}}$

$\boxed{ \sf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}}}$

$\boxed{ \sf{ \: {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)}}$

$\boxed{ \sf{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}$

$\boxed{ \sf{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}$

$\boxed{ \sf{ \: {(x + y)}^{2} = {(x - y)}^{2} + 4xy}}$

$\boxed{ \sf{ \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}$