KnowledgeBoat Logo
|

Mathematics

Simplify :

(x2−y2)3+(y2−z2)3+(z2−x2)3(x−y)3+(y−z)3+(z−x)3\dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3}

Expansions

11 Likes

Answer

If a + b + c = 0; we have :

a3 + b3 + c3 = 3abc.

Since, (x2 - y2) + (y2 - z2) + (z2 - x2) = 0.

∴ (x2−y2)3+(y2−z2)3+(z2−x2)3=3(x2−y2)(y2−z2)(z2−x2)(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2) ………(1)

Also,

(x - y) + (y - z) + (z - x) = 0

∴ (x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)(x - y)^3 + (y- z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) …………..(2)

Dividing equation (1) by (2), we get :

⇒(x2−y2)3+(y2−z2)3+(z2−x2)3(x−y)3+(y−z)3+(z−x)3=3(x2−y2)(y2−z2)(z2−x2)3(x−y)(y−z)(z−x)=3(x−y)(x+y)(y−z)(y+z)(z−x)(z+x)3(x−y)(y−z)(z−x)=(x+y)(y+z)(z+x).\Rightarrow \dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} = \dfrac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)} \\[1em] = \dfrac{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{3(x - y)(y - z)(z - x)} \\[1em] = (x + y)(y + z)(z + x).

Hence, (x2−y2)3+(y2−z2)3+(z2−x2)3(x−y)3+(y−z)3+(z−x)3\dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} = (x + y)(y + z)(z + x).

Answered By

6 Likes

Related Questions