Q1. If the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are equal , then a) a+B+C=0b) a+b(omega)+c(omega)square=0c) a-b+c=0d) none ofd these

Given equation is

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

On simplification, equation changes to

x² - (b + c)x + bc + x² - (c + a)x + ca + x² - (a +b)x + ab = 0

3x² - (2a + 2b + 2c)x + (ab + bc + ca) = 0

3x² - 2(a + b + c)x + (ab + bc + ca) = 0

On comparing the above equation with Ax² + Bx + C = 0,

A = 3, B = -2(a + b + c), C = (ab + bc + ca)

Since roots are equal

Discriminant D = 0

      B² - 4AC = 0

[-2(a + b + c)]² - 4(3)(ab + bc + ca) = 0

4(a + b+ c)² -12(ab + bc + ca) = 0

4(a² + b² + c² + 2ab + 2bc + 2ca) - 12ab -12bc -12ca = 0

4a² + 4b² + 4c² + 8ab + 8bc + 8ca - 12ab -12bc -12ca = 0

4a² + 4b² + 4c² - 4ab -4bc -4ca = 0

4(a² + b² + c² - ab -bc -ca) = 0

Dividing both sides by 4

a² + b² + c² - ab -bc -ca= 0

Hence answer is option D - None of these

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

On simplification, we get

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

Roots are equal implying that D = 0.

i.e 4(a+b+c)^2 -4(3)(ab+bc+ca) = 0

or, 4[a^2 + b^2 +c^2 +2(ab+bc+ca) -3(ab+bc+ca)]=0

or, 4[a^2 + b^2 + c^2 -ab-bc-ca]=0

or, 2[2a^2 + 2b^2 +2c^2 - 2ab -2bc-2ca]=0

or, (a^2 + b^2 -2ab) + (b^2 + c^2 -2bc) + (c^2 + a^2 -2ac)=0

i.e (a-b)^2 + (b-c)^2 + (c-a)^2 = 0

This is true only if all three terms are zero.

Hence a=b=c.

Now option B becomes a( 1+omega + omega(squared) ) = a*0 = 0.

Hence option B is correct.