If ab bc ca 0 then the value of b2 ca c2 ab a2 bc c2 ab a2 bc b2 ca a2 bc b2 ca c2 ab is ........

Quantitative Aptitude (CDS) Simplification

If ab +bc +ca = 0 , then the value of (b² - ca)(c² - ab) + (a² - bc) (c² - ab) + [{ (a² - bc )(b² - ca) } / {(a² - bc )(b² - ca)(c² - ab)}] is

- 1

$\frac{(b^{2}-ca)(c^{2}-ab)+ (a^{2} -bc)(c^{2}-ab)+(a^{2}-bc)(b^{2}-ca)}{(a^{2} -bc)(b^{2}-ca)(c^{2}-ab)}$

ab + bc + ca = 0 .................(i)

b² - ca = b² + ab + bc ................(ii)

= b(b+a+c) = c² - ab

= c² + bc + ca

= c(a+b+c)

Similarly

a² - bc = a(a+b+c)

Equation becomes

{ cb(b+a+c)² + ac( a + b + c)² + ab(a+b+c)² } / {(a² - bc) (b² - ca ) ( c²- ab)}

= (b+a+c)² ( cb + ac + ab) / {(a² - bc) (b² - ca ) ( c²- ab)}

= (b+a+c)²(0) / {(a² - bc) (b² - ca ) ( c²- ab)} = 0

On solving we get

= 0

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