Question

Factorise 2√2a³+16√2b³+c³-12abc

Answer 1

Answer: Factorized form is $(\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)$

Step-by-step explanation:

Since we have given that

$2\sqrt{2}a^3+16\sqrt{2}b^3+c^3-12abc$

We need to factorise the above expression:

As we know the identity of cubes:

$x^3+y^3+x^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Here,

$(\sqrt{2}a)^3+(2\sqrt{2}b)^3+c^3-12abc\\\\=(\sqrt{2}a+2\sqrt{2}b+c)((\sqrt{2}a)^2+(2\sqrt{2}b)^2+c^2-\sqrt{2}a\times 2\sqrt{2}b -2\sqrt{2}bc-\sqrt{2}ac)\\\\=(\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)$

Hence, factorized form is $(\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)$

Answer 2

The answer is in the attachment....

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