(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 Prove A=b=c

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Jul 03, 2024

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Proof: (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 implies a=b=c

In this article, we will prove that the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 implies a=b=c.

Expanding the Equation

First, let's expand the given equation:

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

= x^2 -ax -bx + ab + x^2 -bx -cx + bc + x^2 -cx -ax + ac

= 3x^2 - 3ax - 3bx + 3cx + ab + bc + ac

Rearranging the Terms

Now, let's rearrange the terms to get:

= 3x^2 - x(3a + 3b - 3c) + ab + bc + ac = 0

Dividing by x

Since the equation holds for all values of x, we can divide both sides by x to get:

= 3x - (3a + 3b - 3c) + ab/x + bc/x + ac/x = 0

Taking the Limit as x → ∞

Now, let's take the limit as x approaches infinity:

lim (x → ∞) [3x - (3a + 3b - 3c) + ab/x + bc/x + ac/x] = 0

Since the terms ab/x, bc/x, and ac/x approach 0 as x approaches infinity, we are left with:

= 3x - (3a + 3b - 3c) = 0

Simplifying the Equation

Simplifying the equation, we get:

3a + 3b - 3c = 3x

Equating Coefficients

Since the equation holds for all values of x, we can equate the coefficients of x on both sides:

3a + 3b - 3c = 0

Simplifying Further

Dividing both sides by 3, we get:

a + b - c = 0

Rearranging the Terms

Rearranging the terms, we get:

c = a + b

Using the Original Equation

Now, let's go back to the original equation:

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

Substituting c = a + b, we get:

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

Expanding and Simplifying

Expanding and simplifying the equation, we get:

0 = 0

Conclusion

Since the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 implies c = a + b, and substituting c = a + b back into the original equation results in a true statement, we can conclude that:

a = b = c

Therefore, we have proved that the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 implies a=b=c.