# factorise : a^3-8b^3-64c^3-24abc

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### 1 Answer

We notice that a^3 - 8b^3 - 64c^3 can be get if we'll raise to cube the difference (a - 2b - 4c)^3

We'll raise the trinomial to square:

(a - 2b - 4c)^2 = a^2 + 4b^2 + 16c^2 - 4ab - 8ac + 16bc

Now, we'll multiply the result by (a - 2b - 4c):

(a^2 + 4b^2 + 16c^2 - 4ab - 8ac + 16bc)(a - 2b - 4c) = a^3 - 2a^2*b - 4a^2*c + 4a*b^2 - 8b^3 - 16b^2*c + 16ac^2 - 32bc^2 - 64c^3 - 4a^2*b + 8b^2*a + 16abc - 8a^2*c + 16abc + 32ac^2 + 16abc - 32b^2*c - 64bc^2.

Combining like terms,we'll get:

(a - 2b - 4c)^3 = a^3 - 8b^3 - 64c^3 + 48 abc + ...

Therefore, we'll keep to the left side a^3 - 8b^3 - 64c^3 - 24 abc:

a^3 - 8b^3 - 64c^3 - 24 abc = (a - 2b - 4c)^3 - 72 abc + 6a^2b + 12a^2c - 12ab^2 - 48ac^2 + 48b^2c + 96bc^2

**a^3 - 8b^3 - 64c^3 - 24 abc = (a - 2b - 4c)^3 - 72 abc + 6a^2*(b - 2c) - 12a(b^2 + 2c^2) + 48bc(b + 2c)**