Question

4 x square minus 9 y square whole cube + 9 y square - 16 Z Square whole cube + 16 X square - 4 x square whole cube divided by 2 x minus 3 y whole cube + 3 Y - 47 whole cube + 47 - 2 x whole cube

Answer 1

$\underline{\textsf{To simplify:}}$

$\mathsf{\dfrac{(4x^2-9y^2)^3+(9y^2-16z^2)^3+(16z^2-4x^2)^3}{(2x-3y)^3+(3y-4z)^3+(4z-2x)^3}}$

$\underline{\textsf{Solution:}}$

$\textsf{Formula used:}$

$\boxed{\textsf{If a+b+c=0, then}\;\mathsf{a^3+b^3+c^3=3\,abc}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{(4x^2-9y^2)^3+(9y^2-16z^2)^3+(16z^2-4x^2)^3}{(2x-3y)^3+(3y-4z)^3+(4z-2x)^3}}$

$\textsf{Using the given formula,}$

$\mathsf{=\dfrac{3\,(4x^2-9y^2)(9y^2-16z^2)(16z^2-4x^2)}{3\,(2x-3y)(3y-4z)(4z-2x)}}$

$\mathsf{=\dfrac{((2x)^2-(3y)^2)((3y)^2-(4z)^2)((4z)^2-(2x)^2)}{(2x-3y)(3y-4z)(4z-2x)}}$

$\textsf{using the identity,}$

$\boxed{\mathsf{a^2-b2=(a-b)(a+b)}}$

$\mathsf{=\dfrac{(2x-3y)(2x+3y)(3y-4z)(3y+4z)(4z-2x)(4z+2x))}{(2x-3y)(3y-4z)(4z-2x)}}$

$\mathsf{=(2x+3y)(3y+4z)(4z+2x)}$

$\underline{\textsf{Answer:}}$

$\mathsf{\dfrac{(4x^2-9y^2)^3+(9y^2-16z^2)^3+(16z^2-4x^2)^3}{(2x-3y)^3+(3y-4z)^3+(4z-2x)^3}=(2x+3y)(3y+4z)(4z+2x)}$

Find more:

If cos theta +sec theta = root 3, prove that cos cube theta +sec cube theta = 0

https://brainly.in/question/3077758

Find the value of x3+y3-12xy+64, when x+y=-4

https://brainly.in/question/3807259

If p+q+r=9 then find (3-p)3 +(3-q)3 +(3-r)3

https://brainly.in/question/15155126

Answer 2

Answer:

here u go answer in one page...