Question

a square minus b square whole power 3 plus B squared minus C square whole power 3 + c square minus A square whole power 3 / a minus b whole power 3 plus B minus C whole power 3 +C minus A whole power 3

Answer 1

Answer:

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

Step-by-step explanation:

Given Expression:

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

We need to simplify given expression.

We use the following result,

if x + y + z = 0 then x³ + y³ + z³ = 3xyz

First

x = a² - b²    ,      y = b² - a²     and     z = c² - a²

⇒ x + y + z = a² - b² + b² - c² + c² - a² = 0

So, (a² - b²)³ + (b² - a²)³ + (c² - a²) = 3(a² - b²)(b² - a²)(c² - a²)

Second,

x = a - b     ,      y = b - c      and    z = c - a

⇒ x + y + z = a - b + b - c + c - a = 0

So,  (a - b)³ + (b - c)³ + (c - a)³ = 3(a - b)(b - c)(c - a)

Thus, we get

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

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

Now using, x² - y² = ( x - y )( x + y )

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

=  ( a + b )( b + c )( c + a )

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

Answer 2

Answer:

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

Step-by-step explanation:

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

$We\:know \:that\\</p><p>\boxed {If\:x+y+z=0\:then\:x^{3}+y^{3}+z^{3}=3xyz}$

$Here, x=a^{2}-b^{2},\:y=b^{2}-c^{2},\\c^{2}-a^{2}\\x+y+z\\=a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}\\=0$

$Similarly, a-b+b-c+c-a=0$

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

Therefore,

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

•••♪