Question

Factorize the given equation
 ${a^3} - 8{b^3} - 64{c^3} - 24abc$

Answer 1

Hint: Factorization simply means that it is the process of reducing the brackets of the given quadratic equation, instead of expanding the bracket. Factorization can be also referred to as the reverse function of multiplication. If the above equation is considered, that is a third degree equation, we need to use the following identity to factorize it.
${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z) \times [{(x)^2} + {(y)^2} + {(z)^2} - xy - yz - xz]$
$-8{b}^3$ can be written as $(-2b)^3$
$-64{c^3}$ can be written as $(-4c)^3$

Complete step-by-step solution:
Equation given in the question:
${a^3} - 8{b^3} - 64{c^3} - 24abc$
 We have to factorise the given equation.
For factoring the above equation, it has to be in a certain format, so certain adjustments have to be made that are mentioned in the hint section.
${a^3} - 8{b^3} - 64{c^3} - 24abc$ can be written as ${(a)^3} + {( - 2b)^3} + {( - 4c)^3} - 24abc$
We can also split 24abc in such a way that it includes a product of the coefficients of a2, b2 and c2.
Meaning that, $24abc = 3 \times (a) \times ( - 2b) \times ( - 4c)$
Therefore, on replacing 24abc in the original equation, the equation becomes,
${(a)^3} + {( - 2b)^3} + {( - 4c)^3} - 3 \times (a) \times ( - 2b) \times ( - 4c)$
On comparing with the identity given above in the hint section, we get
${(a)^3} + {( - 2b)^3} + {( - 4c)^3} - 3 \times (a) \times ( - 2b) \times ( - 4c)$
$ = (a - 2b - 4c) \times [{(a)^2} + {( - 2b)^2} + {( - 4c)^2} - a \times ( - 2b) - a \times ( - 4c) - ( - 4c) \times ( - 2b)]$
On simplifying the square bracket, we get,
$ = (a - 2b - 4c) \times [{a^2} + 4{b^2} + 16{c^2} + 2ab + 4ac - 8bc]$
We cannot solve this any further, so this is the final answer.

Note: For different degrees of the equation, there is a different identity. While doing the adjustments, special care should be taken with the sign, most of the mistakes can happen there. Also, try not to solve it more and more and complicate the equation. Keep it simple and stop solving until it becomes complicated.