Find the value of frac{(a-b)^{2}}{(b-c)(c-a)} | vedclass

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(D) Let the given expression be $E = \frac{(a-b)^{2}}{(b-c)(c-a)} + \frac{(b-c)^{2}}{(a-b)(c-a)} + \frac{(a-c)^{2}}{(a-b)(b-c)}$.To add these fraction

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the given expression be $E = \frac{(a-b)^{2}}{(b-c)(c-a)} + \frac{(b-c)^{2}}{(a-b)(c-a)} + \frac{(a-c)^{2}}{(a-b)(b-c)}$.To add these fractions,we take the common denominator as $(a-b)(b-c)(c-a)$.Note that $(c-a) = -(a-c)$,so $(a-c)^2 = (c-a)^2$.Rewriting the expression with the common denominator:$E = \frac{(a-b)^3 + (b-c)^3 + (a-c)^2(a-b)(b-c)/(a-b)(b-c)}{(a-b)(b-c)(c-a)}$Actually,a simpler way is to observe the denominators:$E = \frac{-(a-b)^3 - (b-c)^3 - (c-a)^3}{(a-b)(b-c)(c-a)}$Using the algebraic identity: If $x+y+z = 0$,then $x^3+y^3+z^3 = 3xyz$.Here,let $x = (a-b)$,$y = (b-c)$,and $z = (c-a)$.Then $x+y+z = (a-b) + (b-c) + (c-a) = 0$.Therefore,$(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)$.Substituting this into the expression:$E = \frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 3$.

Find the value of $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(a-c)^{2}}{(a-b)(b-c)}$

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