If a+b+c=0 show that ((a/(b−c))+(b/(c−a))+(c/(a−b)))(((b−c)/a)+((c−a)/b) +((a−b)/c))=9


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Question Number 36019 by math1967 last updated on 27/May/18

$I f a + b + c = 0 s h o w t h a t$ $(\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}) (\frac{b - c}{a} + \frac{c - a}{b} + \frac{a - b}{c}) = 9$
	Answered by ajfour last updated on 27/May/18

	$l e t b = p a, c = q a$ $\Rightarrow a (1 + p + q) = 0 . . . (i)$ $(s h a l l b e u s i n g t h i s i n m a n y$ $s t e p s t h a t f o l l o w . .)$ $l . h . s . = (\frac{1}{p - q} + \frac{p}{q - 1} + \frac{q}{1 - p}) (\frac{p - q}{1} + \frac{q - 1}{p} + \frac{1 - p}{q})$ $= 1 + \frac{(q - 1)}{p (p - q)} + \frac{(1 - p)}{q (p - q)} +$ $\frac{p (p - q)}{q - 1} + 1 + \frac{p (1 - p)}{q (q - 1)} +$ $\frac{q (p - q)}{(1 - p)} + \frac{q (q - 1)}{p (1 - p)} + 1$ $= 3 + \frac{q^{2} - q + p - p^{2}}{p q (p - q)} + \frac{p q (p - q) + p (1 - p)}{q (q - 1)} +$ $\frac{p q (p - q) + q (q - 1)}{p (1 - p)}$ $= 3 + \frac{1 - p - q}{p q} + \frac{p [q p - q^{2} + 1 - p]}{q (q - 1)} +$ $\frac{q [p^{2} - p q + q - 1]}{p (1 - p)}$ $= 3 + \frac{2}{p q} + \frac{p (p - q - 1)}{q} + \frac{q (q - p - 1)}{p}$ $= 3 + \frac{2 + p^{2} (p - q - 1) + q^{2} (q - p - 1)}{p q}$ $= 3 + \frac{2 + p^{3} + q^{3} - p q (p + q) - (p^{2} + q^{2})}{p q}$ $= 3 + \frac{2 - (p^{2} + q^{2} - p q) + p q - 1 + 2 p q}{p q}$ $= 3 + \frac{2 - 1 + 2 p q + p q + p q - 1 + 2 p q}{p q}$ $= 3 + 6 = 9 .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/May/18

	$p r a i s w o r t h y . . .$
	Answered by ajfour last updated on 27/May/18

	$l . h . s . = 3 + Σ (\frac{a}{b - c}) [(\frac{c - a}{b}) + (\frac{a - b}{c})]$ $= 3 + Σ (\frac{a}{b - c}) (\frac{c^{2} - a c + a b - b^{2}}{b c})$ $= 3 + Σ (\frac{a^{2}}{a b c}) (a - b - c)$ $= 3 + Σ (\frac{a^{2}}{a b c}) (2 a) ∵ (- b - c = a)$ $= 3 + 2 Σ \frac{a^{3}}{a b c}$ $a n d a s Σ a^{3} = 3 a b c i f Σ a = 0, s o$ $l . h . s . = 3 + 2 (\frac{3 a b c}{a b c}) = 9 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
	$zLHS 3+(a/(b−c))×(((c−a)/b)+((a−b)/c))+(b/(c−a))×(((b−c)/a)+((a−b)/c))+ (c/(a−b))×(((b−c)/a)+((c−a)/b)) let us calculate (a/(b−c))×(((c−a)/b)+((a−b)/c)) (a/(b−c))×(((c^2 −ac+ab−b^2 )/(bc))) =(a/(b−c))×{((a(b−c)−(b^2 −c^2 ))/(bc))} =(a/(b−c))×{(((b−c)(a−b−c))/(bc))} =(a/(bc))×(a−b−c) =(a/(bc))×2a since a+b+c=0 so (−b−c)=a =2(a^3 /(abc)) other two also give ((2b^3 )/(abc)) and ((2c^3 )/(abc)) so 3+((2a^3 )/(abc))+((2b^3 )/(abc))+((2c^3 )/(abc)) 3+2(((a^3 +b^3 +c^3 )/(abc))) 3+2(((3abc)/(abc))) =3+6 =9 if a+b+c=0 a^3 +b^3 +c^3 =3abc$
	$z L H S$ $3 + \frac{a}{b - c} \times (\frac{c - a}{b} + \frac{a - b}{c}) + \frac{b}{c - a} \times (\frac{b - c}{a} + \frac{a - b}{c}) +$ $\frac{c}{a - b} \times (\frac{b - c}{a} + \frac{c - a}{b})$ $l e t u s c a l c u l a t e$ $\frac{a}{b - c} \times (\frac{c - a}{b} + \frac{a - b}{c})$ $\frac{a}{b - c} \times (\frac{c^{2} - a c + a b - b^{2}}{b c})$ $= \frac{a}{b - c} \times {\frac{a (b - c) - (b^{2} - c^{2})}{b c}}$ $= \frac{a}{b - c} \times {\frac{(b - c) (a - b - c)}{b c}}$ $= \frac{a}{b c} \times (a - b - c)$ $= \frac{a}{b c} \times 2 a s i n c e a + b + c = 0 s o (- b - c) = a$ $= 2 \frac{a^{3}}{a b c}$ $o t h e r t w o a l s o g i v e \frac{2 b^{3}}{a b c} a n d \frac{2 c^{3}}{a b c}$ $s o$ $3 + \frac{2 a^{3}}{a b c} + \frac{2 b^{3}}{a b c} + \frac{2 c^{3}}{a b c}$ $3 + 2 (\frac{a^{3} + b^{3} + c^{3}}{a b c})$ $3 + 2 (\frac{3 a b c}{a b c})$ $= 3 + 6$ $= 9$ $i f a + b + c = 0$ $a^{3} + b^{3} + c^{3} = 3 a b c$
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