Question

please help ... I have asked this question earlier too...but no correct answers....Q no. 4​

Answer 1

first of all take out the factor theorem of all three values then when you get your answer substitute the value in all then you will get your answer

Answer 2

Given :

a (b² - c²) + b (c² - a²) + c (a² - b²)

Let f(a - b) = a (b² - c²) + b (c² - a²) + c (a² - b²)

⇒ f(a - b) =ab² - ac² + bc² - a²b + c (a + b)(a - b)

⇒ f(a - b) = ab² - a²b + bc²- ac² + c(a + b)(a - b)

⇒ f(a - b) = ab(b - a) + c²(b - a) + c(a + b)(a - b)

⇒ f(a - b) = (b - a)(ab + c² + c(a + b) )

When a - b is 0 , f(a - b) = 0 and hence a - b is a factor of the function ( using the Factor theorem ) .

Let f(b - c) = a (b² - c²) + b (c² - a²) + c (a² - b²)

⇒ f(b - c) = a(b + c)(b - c) + bc² - b²c - a²b + a²c

⇒ f(b - c) = a(b + c)(b - c) - bc(b - c) - a²(b - c)

⇒ f( b - c) = (b - c)( a(b + c) - bc - a² )

When b - c is 0 , f(b - c) = 0 and hence b - c is a factor of the function ( using the Factor theorem ) .

Let f(c - a) = a (b² - c²) + b (c² - a²) + c (a² - b²)

⇒ f(c - a) = ab² - ac² + b(c + a)(c - a) + a²c - b²c

⇒ f(c - a) = -b²(c - a) - ac(c - a) + bc(c + a)(c - a)

⇒ f(c - a) = (c - a)( -b²- ac + bc(c + a ) )

When c - a is 0 , f(c - a) = 0 and hence c - a is a factor of the function ( using the Factor theorem ) .