Question

(a-b)^2/(b-c)(c-a)+(b-c)^2/(a-b)(c-a)+(c-a)^2/(a-b)(b-c)​

Answer 1

Answer:

3

hope it will help u...

Answer 2

(a-b)²/(b-c)(c-a) + (b-c)²/(a-b)(c-a)+(c-a)²/(a-b)(b-c)​ = 3

Given:

(a-b)²/(b-c)(c-a) + (b-c)²/(a-b)(c-a)+(c-a)²/(a-b)(b-c)​

To find:

Find the value of (a-b)²/(b-c)(c-a) + (b-c)²/(a-b)(c-a)+(c-a)²/(a-b)(b-c)​

Solution:

Given (a-b)²/(b-c)(c-a) + (b-c)²/(a-b)(c-a)+(c-a)²/(a-b)(b-c)​  

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

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

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

From algebraic identities

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

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

Therefore,

(a-b)²/(b-c)(c-a) + (b-c)²/(a-b)(c-a)+(c-a)²/(a-b)(b-c)​ = 3

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