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The roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c)=0 are equal then prove that a+bw+cw^2=0 w is imaginary cube root of unity

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one year ago

```							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
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one year ago
```							once again, aruns answer is incomplete and slightly wrong.correct method: 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 = 0NOW COMES THE IMPORTANT PART:we need to use the following identity: a² + b² + c² - ab - bc - ca = (a+bw+cw^2)(a+bw^2+cw)so that (a+bw+cw^2)(a+bw^2+cw)= 0which means either a+bw+cw^2= 0 or a+bw^2+cw= 0if a+bw+cw^2= 0, we jave proved the desired result.if a+bw^2+cw= 0, then we can take conjugate on both sidesso a+b*conj(w^2)+c*conj(w)= 0but we know that conj(w)= w^2 and conj(w^2)= wso a+bw+cw^2= 0hence proved.kindly approve yar :)
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one year ago
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