Question

(a²-b²)³ + (b²-c²)³ + (c²-a²)³/(a-b)³ + (b-c)³ + (c-a)³​

Answer 1

Answer:

(a + b)(b + c)(c + a)

Step-by-step explanation:

Given :

To simplify the expression,

$\frac{(a^2-b^2)^3 +(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3 +(b-c)^3+(c-a)^3}$

Solution :

We know that,

$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

Simplifying the numerator,

As,

$x =(a^2 - b^2)$

$y=(b^2-c^2)$

$z=(c^2-a^2)$

So,

$x+y+z = (a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$,

⇒ $x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)$

⇒ $x^3 + y^3 + z^3 - 3xyz = 0$

⇒ $x^3 + y^3 + z^3 = 3xyz$

⇒ $(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$

We know that,

$x^2-y^2=(x+y)(x-y)$

So,

$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)=3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$

Simplifying the denominator,

As,

$x=(a-b)^3$

$y=(b-c)^3$

$z=(c-a)^3$

So,

$x+y+z=(a-b)+(b-c)+(c-a)=0$

⇒ $x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)$

⇒ $x^3 + y^3 + z^3 - 3xyz = 0$

⇒ $x^3 + y^3 + z^3 = 3xyz$

⇒ $(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$

By substituting the values of numerator & denominator,

We get,

⇒ $\frac{(a^2-b^2)^3 +(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3 +(b-c)^3+(c-a)^3}=\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}=(a+b)(b+c)(c+a)$