If x=a(b-c), y=b(c-a) and z=c(a-b), then left | vedclass

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(A) Given: $x=a(b-c), y=b(c-a), z=c(a-b)$.Dividing by $a, b, c$ respectively,we get:$\frac{x}{a} = b-c, \frac{y}{b} = c-a, \frac{z}{c} = a-b$.Let $p =

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$\frac{3xyz}{abc}$

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(A) Given: $x=a(b-c), y=b(c-a), z=c(a-b)$.Dividing by $a, b, c$ respectively,we get:$\frac{x}{a} = b-c, \frac{y}{b} = c-a, \frac{z}{c} = a-b$.Let $p = \frac{x}{a}, q = \frac{y}{b}, r = \frac{z}{c}$.Then $p+q+r = (b-c) + (c-a) + (a-b) = 0$.We know the algebraic identity: If $p+q+r=0$,then $p^3+q^3+r^3 = 3pqr$.Substituting the values back:$\left(\frac{x}{a}\right)^3 + \left(\frac{y}{b}\right)^3 + \left(\frac{z}{c}\right)^3 = 3 \left(\frac{x}{a}\right) \left(\frac{y}{b}\right) \left(\frac{z}{c}\right) = \frac{3xyz}{abc}$.

If $x=a(b-c), y=b(c-a)$ and $z=c(a-b),$ then $\left(\frac{x}{a}\right)^{3}+\left(\frac{y}{b}\right)^{3}+\left(\frac{z}{c}\right)^{3}=?$

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