Question

(a - b)(a + b) + (6-c) (b + c)+(c - a) (c + a) = 0​

Answer 1

Step-by-step explanation:

Given, 2a+3b+6c=0

Let f′(x)=ax2+bx+c

f(x)=3ax3+2bx2+cx+d

⇒f(x)=62ax3+3bx2+6cx+6d

Now, f(1)=62a+3b+6c+6d=66d=d

and f(0)=66d=d

∴f(0)=f(1)

⇒f′(x) will vanish atleast once between 0 and 1. (by Rolle's theorem)

Therefore, one of the roots of the equation ax2+bx+c=0 lies between 0 and 1.

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Answer 2

Step-by-step explanation:

Given:-

(a - b)(a + b) + (6-c) (b + c)+(c - a) (c + a) = 0

Correct question:-

Show that

(a - b)(a + b) + (b-c) (b + c)+(c - a) (c + a) = 0

Taking LHS

(a - b)(a + b) + (b-c) (b + c)+(c - a) (c + a)

=>a²-b²+b²-c²+c²-a²

(since (x+y)(x-y)=x²-y²)

=>(a²-a²)+(b²-b²)+(c²-c²)

=>0+0+0

=>0

=RHS

LHS=RHS

(a - b)(a + b) + (b-c) (b + c)+(c - a) (c + a) = 0