Question

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Plz Answer fast

(xa/xb)a²+2ab+b²×(xb/xc)×b²+2bc+c²×(xa/xc)c²+2ca+a² = 1

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Answer 1

Explanation:

Root of quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal

Means D = b² - 4ac = 0 for this equation,

first we should rearrange the equation ,

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

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

⇒3x² - 2(a + b + c)x + (ab + bc + ca)

D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0

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

⇒ a² + b² + c² - ab - bc - ca = 0

⇒2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

⇒(a - b)² + (b - c)² + (c - a)² = 0

This is possible only when , a = b = c

Hence, proved , if roots of given equation are equal then, a = b = c

Answer 2

Answer:

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