Question

Factorise (x^15 - 64y^3)​

Answer 1

Answer:

using identity

a3-b3 (a-b)(a2+ab+b2)

x-4y)(x2+4yx+16y2)

Answer 2

Answer:

$x^{15}-64y^{3}=(x^{5}-4y )[x^{10}+4x^{5}y+16y^{2}]$

Step-by-step explanation:

In the question,

We have,

$x^{15}-64y^{3}$

We know from the property of Difference of the cubes that,

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

So,

Using the same formula of the Difference of cubes for the expansion of the equation in the given question, we get,

On factorising,

$x^{15}-64y^{3}=(x^{5})^{3} -(4y)^{3}  =(x^{5}-4y )[x^{10}+4x^{5}y+16y^{2}    ]$

Therefore, we get the final factorised equation as,

$x^{15}-64y^{3}=(x^{5}-4y )[x^{10}+4x^{5}y+16y^{2}]$

It can not be factorised further therefore this is the final factorised equation.