Question

what is the factorization of 216x^12 -64​

Answer 1

The factorization of given expression is:

$216x^{12}-64 = 8\left(3x^4-2\right)\left(9x^8+6x^4+4\right)$

Solution:

Given that we have to factorize the given expression

Given expression is:

$216x^{12}-64$

Let us factorize the expression

$\text{ Rewrite } 64 \text{ as } 8 \times 8$

$\text{Rewrite } 216 \text{ as } 8 \times 27$

Thus the given expression becomes,

$216x^{12} - 64 = (8 \times 27)(x^{12}) - (8 \times 8)\\\\\text{Factor out common term 8 }$

$216x^{12} -64= 8(27x^{12}-8)$

$\mathrm{Rewrite\:}27x^{12}-8\mathrm{\:as\:}\left(3x^4\right)^3-2^3$

$8(27x^{12}-8) = 8(\left(3x^4\right)^3-2^3)$

Let us apply the difference of cubes formula

$x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)$

$\left(3x^4\right)^3-2^3=\left(3x^4-2\right)\left(3^2x^8+2\cdot \:3x^4+2^2\right)$

Therefore,

$8(27x^{12}-8) = 8\left(3x^4-2\right)\left(3^2x^8+2\cdot \:3x^4+2^2\right)\\\\8(27x^{12}-8) = 8\left(3x^4-2\right)\left(9x^8+6x^4+4\right)$

Thus factorization of given expression is:

$216x^{12}-64 = 8\left(3x^4-2\right)\left(9x^8+6x^4+4\right)$