Question

25. Solve this problem
Please give verified answer with step by step solution​

Answer 1

$\red{\bf{(a+b)(b+c)(c+a)}}$

Step-by-step explanation:

To Solve:

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

Solution:

It's seem as a difficult question but we can solve it easily by using this formula.

If x + y + z = 0 , Therefore x³ + y³ + z³ = 3xyz

Now, Checking Formula is applicable for this question...

let, x = (a - b) , y = (b - c) and z = (c - a)

So, x + y + z = a - b + b - c + c - a = 0

Similarly, x = (a² - b²) , y = (b² - c²) and z = (c² - a²)

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

I get, Formula is applicable....

$\sf\:\dfrac{(a^{2}\:-\:b^{2})^{3}\:+\:(b^{2}\:-\:c^{2}\:)^{3}\:+\:(c^{2}\:-\:a^{2})^{3}}{(a-b)^{3}\:+\:(b-c)^{3}\:+\:(c-a)^{3}}\\\\\implies\sf\:\dfrac{3\:(a^{2}-b^{2})\:(b^{2}\:-\:c^{2})\:(c^{2}-a^{2}}{3(a-b)\:(b-c)\:(c-a)}\\\\\sf\:Using\:x^{2}\:-\:y^{2}\:=\:(x+y)\:(x-y)\:we\:get\\\\\sf\:\dfrac{3(a+b)\:(b+c)\:(c+a)\:(a-b)\:(b-c)\:(c-a)}{3(a-b)\:(b-c)\:(c-a)}\\\\\implies\sf\:\dfrac{\cancel{3}(a+b)\:(b+c)\:(c+a)\:\cancel{(a-b)\:(b-c)\:(c-a)}}{\cancel{3}\cancel{(a-b)\:(b-c)\:(c-a)}}\\\\\implies\sf \:(a+b)\:(b+c)\:(c+a)$

It is easy, Isn't ?

Answer 2

Answer:

nhi janta.

mark me an brainlist