Question

a cube +8b cube +64c cube -24abc

Answer 1

${a}^{3} + 8 {b}^{3} + 64 {c}^{3} - 24abc \\ = {a}^{3} + (2 {b})^{3} + (4 {c})^{3} - 3 \times a \times 2b \times 4c \\ (a + 2b + 4c) ( {a}^{2} + {2b}^{2} + {4c}^{2} - a(2b) - (2b)(4c) - (4c)a)\\ = (a + 2b + 4c) ( {a}^{2} + 4{b}^{2} + 16{c}^{2} - 2ab - 8bc- 4ca)$

Answer 2

Here is your solution

we know that:-

$a {}^{3} + b {}^{3} + c {}^{3} - 3ab =(a + b + c)(a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca)$

Now,

$a {}^{3} + 8b {}^{3} + 64c {}^{3} - 24abc = (a + 2b + 4c)(a {}^{2} + 4b {}^{2} + 16c {}^{2} - 2ab - 8bc - 4ac)$